Strand 2 of 5. 6 th Year Maths Ordinary Level. Topics: Trigonometry Co-ordinate Geometry of the Line Co-ordinate Geometry of the Circle Geometry

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1 6 th Year Maths Ordinary Level Strand 2 of 5 Topics: Trigonometry Co-ordinate Geometry of the Line Co-ordinate Geometry of the Circle Geometry No part of this publication may be copied, reproduced or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without prior written permission from The Dublin School of Grinds. (Notes reference: 6-mat-o-Strand 2).

2 MATHS PAPER 2 BLOCK COURSES One last drive for Maths Paper 2... The Dublin School of Grinds is running four-hour block courses on Sunday 11th June. These courses are designed for those students who feel that they could use an extra push after Maths Paper 1. This exam-focused and structured revision environment could be all you need to help increase your Maths Paper 2 result. Note: At these courses our teachers will predict what questions are most likely to appear on your Maths Paper 2 exam. All these questions will be covered in detail and our teachers will provide you with the techniques to answer each question. FEES: 160 PER COURSE To book your place, call us on: or visit: Maths Paper 2 Block Courses Timetable 6th Year SUBJECT LEVEL DATE TIME Maths H Sunday 11th June 9am - 1pm Maths O Sunday 11th June 9am - 1pm DATE TIME Sunday 11th June 9am - 1pm 3rd Year SUBJECT Maths LEVEL H Note: All courses will take place in Stillorgan Plaza, Lower Kilmacud Road, Stillorgan, Co. Dublin. DSOG January 12pg A4 Brochure.indd 4 20/11/ :32

3 Strand 2 is worth 20% to 31% of The Leaving Cert. Contents: Trigonometry 1. Types of triangles Pythagoras Theorem Sin, cos and tan Using the calculator for sin, cos and tan Solving right angled triangles: an unknown side Solving right angled triangles: an unknown angle An angle of elevation An angle of depression Compass directions Area of a triangle The Sine Rule The Cosine Rule The sector of a circle Practical problems Past and probable exam questions Solutions to Trigonometry Co-ordinate Geometry of the Line 1. Finding the distance between two points (using the distance formula) Finding the midpoint of a line segment (using the midpoint formula) Finding the endpoint of a line segment when given the midpoint Finding the slope of a line Finding slopes of parallel and perpendicular lines Finding the equation of a line (using the equation formula) Checking if a point is on a line Plotting lines on a co-ordinate diagram Finding the point of intersection of two lines Finding the area of a triangle using the formula Using translations Past and probable exam questions Solutions to Co-ordinate geometry of the line The Dublin School of Grinds Page 1

4 Co-ordinate geometry of the circle 1. Equation of a circle with centre (0, 0) Point inside, on or outside a circle with centre (0, 0) The Equation of a circle with the centre not at (0, 0) Point in, on or outside a circle with centre not (0, 0) Point of intersection between a circle and a line Finding the equation of a tangent Finding the points of intersection with the axes Past and probable exam questions Solutions to Co-ordinate geometry of the circle Geometry 1. Definitions Constructions Theorems Basic rules of angles and triangles Quadrilaterals Basic rules of circles Transformations Enlargements Past and probable exam questions Solutions to Geometry The Dublin School of Grinds Page 2

5 Trigonometry is worth 8% to 12% of the Leaving Cert. It appears on Paper 2. Trigonometry 1. Types of triangles The Examiner requires you to know the names of the different types of triangles. There are 3 types: Type 1: When all sides and angles are equal it is called an equilateral triangle or equiangular triangle. Type 2: When 2 sides and 2 angles are equal it is called an isosceles triangle : Type 3: When all sides and all angles are different it is called a scalene triangle : The Dublin School of Grinds Page 3

6 2. Pythagoras Theorem You need to be able to use Pythagoras theorem in the Leaving Cert exam to find an unknown side of a right-angled triangle. On page 16 of the log tables you will see the following: c 2 = a 2 + b 2 Example 1 Find the length of the side marked x in the following right-angled triangle. Solution a = 5 b = 12 c = x x 2 = x 2 = 169 x = 169 = 13cm Question 2.1 What is the length of the side PR? The Dublin School of Grinds Page 4

7 Question 2.2 In the rectangle PQRS, PS = 16cm and RS = 8cm, find the length of the diagonal PR, correct to one decimal place. The Dublin School of Grinds Page 5

8 3. Sin, cos and tan The Leaving Cert Syllabus requires you to know how to find the sin, cos or tan of angles. sin A = cos A = tan A = OPPOSITE SIDE HYPOTENUSE ADJACENT SIDE HYPOTENUSE OPPOSITE SIDE ADJACENT SIDE => S = O H => C = A H => T = O A The way to remember this is SOH CAH TOA or Silly Old Harry Caught A Herring Trawling Off America. Example 1 Write down the Sin, Cos and Tan of the angle marked with a capital letter in each of the following triangle: Hypotenuse = 17 Opposite = 8 Adjacent = 15 sina = 8 17 cosa = tana = 8 15 Question 3.1 Write down the sin, cos and tan of the angle marked A. The Dublin School of Grinds Page 6

9 Question 3.2 If sina = 5, find tana. 13 The Dublin School of Grinds Page 7

10 4. Using the calculator for sin, cos and tan Example 1 Use your calculator to evaluate each of the following, correct to 4 decimal places. i) Sin36 = ii) Cos = iii) Tan48 17 = Solution i) ii) Sin36 = To input the seconds press the buttons shown below: iii) Sharp calculators cos 27 D 0 M S 31 = Casio calculators cos = Use the same buttons as above: Cos = Tan48 17 = Question 4.1 Use your calculator to evaluate each of the following, correct to 4 decimal places. i) sin16 ii) tan48 16 iii) cos56 40 The Dublin School of Grinds Page 8

11 Example 2 Find the value of the each of the angles to the nearest degree: i) tana = 0.42 ii) Solution i) cosθ = 3 2 ii) tana = 0.42 A = tan A = 23 cosθ = 3 2 θ = cos 1 ( 3 2 ) θ = 30 Question 4.2 Find the value of the angle in each of the following, giving your answer correct to the nearest degree: i) CosA = ii) sina = 1 2 The Dublin School of Grinds Page 9

12 5. Solving right angled triangles: an unknown side The Examiner can ask you to use sin, cos or tan to find an unknown side of a right angled triangle. Example 1 PQR is a triangle in which PQ = 8cm and RPQ = 30, find QR, correct to two decimal places. Solution First write the information you are given on the triangle: We know the hypotenuse and we want to find the opposite side so we will use sin. sin30 = opposite hypotenuse sin30 = x 8 8sin30 = x 4cm = x Question 5.1 Using the information given in the diagram below, find the height of the flagpole correct to two decimal places. The Dublin School of Grinds Page 10

13 6. Solving right angled triangles: an unknown angle Example 1 In the triangle PQR, PR =9m, QR = 13m and PRQ = 90. Find PQR, correct to the nearest degree. Solution First, fill the information in on the diagram: Since you have opposite and adjacent we will use tan. tana = Opposite Adjacent tana = 9 13 A = tan 1 ( 9 13 ) => A = 35 Question 6.1 In the diagram below, find the measure of the angle marked θ correct to 2 decimal places The Dublin School of Grinds Page 11

14 7. An angle of elevation Sometimes when the Examiner gives you a problem he will use the term angle of elevation. In the Leaving Cert exam you can be asked to describe how to find the angle of elevation. Learn these steps, they are free marks if you are asked. 1. Tilt the clinometer as you look through the eyepiece so that the highest point on the top of the building is visible. 2. Read the angle of elevation of this highest point to the nearest degree. Example 1 The Sun casts an 8m shadow from a tree. The Sun is at an angle of What is the height of the tree correct to one decimal place? Solution The tree is the opposite side and the shadow is adjacent so we will use tan. tana = opposite adjacent tan35 = x 8 8tan35 = x 5.6m = x The Dublin School of Grinds Page 12

15 Question 7.1 A wall casts a shadow on the ground. The distance from the top of the wall to the end of the shadow is 13m. The angle of elevation of the top of the wall to the end of the shadow is Find the height of the wall correct to one decimal place. The Dublin School of Grinds Page 13

16 8. An angle of depression Another term the Examiner likes to use is angle of depression. This occurs when you are looking down. (If you re depressed you re looking down). Question 8.1 From the top of the vertical cliff, the angle of depression of a yacht at sea is found to be The yacht is 65 metres from the base of the cliff. Calculate the height of the cliff, correct to the nearest metre. The Dublin School of Grinds Page 14

17 9. Compass directions In the Leaving Cert exam you can be given a question with compass directions so you must learn how to follow them. Example 1 Indicate the following on a diagram: i) N40 0 W ii) N30 0 E iii) S70 0 E iv) S17 0 W Solution To find the direction follow these steps: 1. Go to the direction of the first letter. 2. Move through the angle given towards the direction of the second letter. N40 W N N30 E W 70 E 17 S70 E S17 W S Question 9.1 Indicate the following directions on the diagram given: i) N50 E ii) S10 W iii) S60 E The Dublin School of Grinds Page 15

18 Question 9.2 Two ships a and b leave a harbour x, a sails in the direction S58 E and b sails in the direction S32 W. a sails a distance of 21km and b sails a distance of 21km. How far are the ships apart to the nearest kilometre? The Dublin School of Grinds Page 16

19 10. Area of a triangle The Examiner can ask you to find the area of a triangle using trigonometry. If we know two sides of a triangle and the angle between the two sides, then the area of the triangle can be found. On page 16 of the log tables you will see this: Example 1 In the triangle PQR, XY = 5cm, YZ = 4cm and XYZ = 48. Find the area of the XYZ, correct to one decimal place. Solution Area = 1 ab sin C 2 = 1 2 (4)(5)sin48 = 7.4cm 2 Question 10.1 Find the area of the given triangle correct to one decimal place. Solution The Dublin School of Grinds Page 17

20 You may be asked to find the area of other shapes where you will divide it up into triangles. Question 10.2 Find the area of the quadrilateral WXYZ. Give your answer correct to the nearest integer. Finding a side or an angle using the area In the Leaving Cert exam you may be asked to find an unknown side or angle when you are given the area. Example 2 In the triangle ABC, the area is 48cm 2, BC = 6cm and ABC =50 0. Find AB correct to two decimal places. Solution Let s draw a sketch: Area = 1 ab sin C 2 48 = 1 2 (6)(x)sin50 48 = (x)3sin sin50 = x 20.89cm = x Question 10.3 The area of the triangle shown is 30 square units. Find the length of the side marked x, correct to one decimal place. The Dublin School of Grinds Page 18

21 Example 3 The area of the triangle below is 29cm 2, find the size of the angle A to the nearest degree. Solution Area = 1 ab sin C 2 29 = 1 (6)(10) sin A 2 29 = 30 sin A = sina sina = A = sin 1 ( ) => A = 75 The Dublin School of Grinds Page 19

22 Question 10.4 The area of the triangle given in the diagram is 48cm 2. Find the size of the angle marked θ. The Dublin School of Grinds Page 20

23 11. The Sine Rule The Leaving Cert syllabus requires you to know how to use the sine rule to find a missing side or a missing angle in a nonright angled triangle. On page 16 of the log tables you will see the following: NOTE: small letters = sides CAPITAL LETTERS = ANGLES NOTE: Your teacher may tell you to use this formula upside down. DON T. Example 1 Find the length of the side marked x correct to one decimal place. Solution a sina = b sinb x sin85 = 15 sin45 => x = 15sin85 sin45 = 21.1 The Dublin School of Grinds Page 21

24 Question 11.1 In the triangle ABC, AC = 16m, ABC = 48 and BAC 50. Find BC, correct to the nearest metre. Question 11.2 Find the measure of the angle C in the triangle shown, correct to two decimal places. The Dublin School of Grinds Page 22

25 12. The Cosine Rule When Pythagoras/sin/cos/tan/sine rule don t work your last resort is the Cosine Rule. On page 16 of the log tables you will see the following: NOTE: Whatever side you use for a, you must use the angle across from it for A. Example 1 Using the Cosine Rule find the length of the side marked x, correct to two decimal places. Solution a 2 = b 2 + c 2 2bc cos A x 2 = (8) 2 + (6) 2 2(8)(6) cos 50 = ( ) = = => x = = 6.19cm The Dublin School of Grinds Page 23

26 Question 12.1 Using the Cosine Rule find the length of the side marked a, correct to two decimal places. When you need to use the Cosine Rule to find an angle we can rearrange it to look like this: cosa = b2 + c 2 a 2 2bc NOTE: This formula is not in the log tables and you must learn it off by heart Example 2 In the triangle AB = 5cm, BC = 4cm and AC = 8cm. Find ABC, correct to the nearest degree. Solution cosx = b2 + c 2 a 2 2bc cosx = (5)(4) = 40 = => X = cos 1 ( ) = 125 Note: In the Question the Examiner can use sly tactics by using the letters A, B and C just like in the formula given in the formula and tables booklet. The Dublin School of Grinds Page 24

27 Question 12.2 Use the Cosine Rule to find the size of the angle marked θ, correct to two decimal places. The Dublin School of Grinds Page 25

28 13. The sector of a circle The Leaving Cert syllabus requires you to know how to find the length of an arc and the area of a sector. The formulae are found on page 9 of the log tables.: NOTE: A sector is like a pizza slice. An arc is like a pizza crust. Example 1 The given circle has radius 18cm and AOB = 75. Taking π = 22, find 7 i) The length of the minor arc AB. Correct to one decimal place. ii) The area of the sector AOB, correct to one decimal place. Solution i) ii) l = 2πr ( θ 360 ) = (2) ( ) (18) ( ) = 23.6cm A = πr 2 ( θ 360 ) = ( 22 7 ) (18)2 ( ) = 212.1cm 2 Question 13.1 Find the area and the length of the arc of the sector in the diagram, correct to one decimal place (taking π = 3.14). The Dublin School of Grinds Page 26

29 Question 13.2 The given sector has a radius of 16cm. The angle in the sector is Find correct to the nearest whole number in each case (taking π = 3.14). i) The area of the sector OAB. ii) The area of the triangle OAB. iii) The area of the shaded region. The Dublin School of Grinds Page 27

30 14. Practical problems The Examiner can give you questions in the Leaving Cert exam where you must use a combination of the rules we have seen. Example 1 A path goes around a forest so that it forms the three sides of a triangle. The lengths of two sides are 550m and 290 m. These two sides meet at an angle of 115. A diagram is shown below. (a) (b) Calculate the length of the third side of the triangle. Give your answer correct to the nearest metre. Calculate the area enclosed by the path that goes around the forest. Inside the forest a second path forms the three sides of another triangle named ABC. Angle BAC is 53, AC is 180 m and BC is 230 m. (c) Calculate the size of angle ABC, correct to the nearest degree. Solution a) a 2 = b 2 + c 2 2bc cos A x 2 = (290) 2 + (550) 2 2(290)(550) cos 115 = 84, , ,000( ) = 386, ,809.4 = 521,409.4 => x = 521,409.4 = 722m b) Area = 1 ab sin C 2 = 1 2 (290)(550)sin115 c) = 72,278m 2 a sina = b sinb 180 siny = 230 sin53 180sin53 = 230siny 180sin53 = siny 230 sin 1 ( 180sin53 ) = y = y The Dublin School of Grinds Page 28

31 Question 14.1 A cross-country running course is given in the diagram below. Runners start and finish at point O. O Not to scale C 1500 m 500 m B 800 m 110 A (a) (b) Show that the distance CA is 943 m correct to the nearest metre. Show that angle BCA is 58 correct to the nearest degree. (c) (i) Calculate the angle CAO. (ii) Calculate the distance CO. (d) Calculate the area enclosed by the course OABC. Solution The Dublin School of Grinds Page 29

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33 15. Past and probable exam questions Question 1 The Dublin School of Grinds Page 31

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35 Question 2 The Dublin School of Grinds Page 33

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37 Question 3 Find the area of the triangle BCD. The Dublin School of Grinds Page 35

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46 Question 8 b and c are two airports as shown. When airport b is viewed from a, abc = 36. When airport c is viewed from a, acb = 44. It takes a plane 25 minutes travelling at a speed of 384km/h to go from airport b to airport c. Find (i) the distance between both airports, i.e. bc, (ii) the distance airport c is from point a, i.e. ac, Correct to the nearest km. The Dublin School of Grinds Page 44

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53 16. Solutions to Trigonometry Question 2.1 a = 9 b = x c = 15 Question x 2 = x 2 = 225 x 2 = x 2 = 144 x = 144 x = 12 a = 16 b = 8 c = x Question 3.1 x 2 = x 2 = 320 x = 320 = 17.9cm Hypotenuse = 5 Opposite = 3 Adjacent = 4 SinA = 3 5 CosA = 4 5 Question 3.2 Opposite = 5 Hypotenuse = 13 TanA = 3 4 sina = 5 13 a = 5 b = x c = x 2 = x 2 = 169 x 2 = x 2 = 144 x = 144 x = 12 => Adjacent = 12 => TanA = 5 12 The Dublin School of Grinds Page 51

54 Question 4.1 i) sin16 = ii) tan48 16 = iii) cos56 40 = Question 4.2 i) ii) cosa = A = cos A = 41 sina = 1 2 A = sin 1 ( 1 2 ) A = 45 Question 5.1 sin30 = Opposite Hypotenuse sin28 = x 20 20sin28 = x 9.39m = x Question 6.1 sinθ = Opposite Hypotenuse sinθ = 4 41 θ = sin 1 ( 4 41 ) θ = Question 7.1 sina = Opposite Hypotenuse sin25 = x 13 13sin25 = x 5.5m = x => height = 5.5m The Dublin School of Grinds Page 52

55 Question 8.1 tana = Opposite Adjacent tan47 = x 65 65tan47 = x 70m = x => height = 70m Question 9.1 Question = 90 0 (the ships leave at an angle of 90 0 so we can use Pythagoras Theorem) x 2 = x 2 = 882 x = 882 x = => x = 30km Question 10.1 Area = 1 ab sin C 2 = 1 2 (8)(5)sin65 = 18.1m 2 The Dublin School of Grinds Page 53

56 Question 10.2 To find the area of the quadrilateral split it into two triangles WYZ and WXY. Area of WYZ = 1 ab sin C 2 = 1 2 (14)(24)sin45 = 84 2units 2 Area of WXY = 1 ab sin C 2 = 1 2 (24)(20)sin68 = units 2 Total Area of WXYZ = = 341units 2 Question 10.3 Area = 1 ab sin C 2 30 = 1 2 (12)(x)sin37 30 = (x)6sin sin37 = x 8.3cm = x Question 10.4 Area = 1 ab sin C 2 48 = 1 (24)(8) sin θ 2 48 = 96 sin θ = sinθ sinθ = θ = sin 1 ( ) => θ = 30 Question 11.1 a sina = b sinb x sin50 = 16 sin48 => x = 16sin50 sin48 = 16m The Dublin School of Grinds Page 54

57 Question 11.2 a sina = b sinb 6 sinc = 12 sin61 6sin61 = 12sinC => 6sin61 12 = sinc sin 1 ( 6sin61 ) = C 12 => = C Question 12.1 a 2 = b 2 + c 2 2bc cos A a 2 = (4) 2 + (6) 2 2(4)(6) cos 45 = ( ) = = => a = = 4.25m Question 12.2 cosθ = b2 + c 2 a 2 2bc cosθ = (9)(9) = = => θ = cos 1 ( ) = Question 13.1 i) A = πr 2 ( θ 360 ) = (3.14)(11) 2 ( ) = 126.6cm 2 ii) l = 2πr ( θ 360 ) = (2)(3.14)(11) ( ) = 23.0cm The Dublin School of Grinds Page 55

58 Question 13.2 i) Area of sector OAB = πr 2 ( θ 360 ) = (3.14)(16) 2 ( ) = 179cm 2 ii) Area of triangle OAB = 1 ab sin C 2 = 1 2 (16)(16)sin80 iii) = 126cm 2 Area of shaded region = Area of sector OAB Area of triangle OAB = = 53cm 2 The Dublin School of Grinds Page 56

59 Question 14.1 a) We can use Pythagoras Theorem to find the distance of CA Let CA = x x 2 = x 2 = 890,000 x = 890,000 x = 943m b) The angel BCA is marked with the symbol θ. c) i) sinθ = Opposite Hypotenuse = => θ = sin 1 ( ) = 58 CAB = = 32 CA0 = 110 CAB = = 78 ii) We can look at the triangle COB to find the length of CO. x a 2 = b 2 + c 2 2bc cos A x 2 = (943) 2 + (1500) 2 2(943)(1500) cos 78 = 3,139, , = 2,551, => x = 2,551, = 1597m d) To find the area enclosed by OABC we can find the area of triangles ABC and ACO. The Dublin School of Grinds Page 57

60 Area of ABC = 1 (800)(943) sin 32 2 = 199,886m 2 Area of ACO = 1 (943)(1500) sin 78 2 = 691,795m 2 => Area = 199, ,795 = 891,681m 2 The Dublin School of Grinds Page 58

61 Past and probable exam questions solutions Question 1 a) cm 22cm 20cm 18cm b) a sina = b sinb 25 sinx = 22 sin60 25sin60 = 22sinX => 25sin60 22 = sinx sin 1 ( 25sin60 ) = X 22 => 79.8 = X => CFA = 79.8 c) d) Use the Cosine Rule to find the length of the strap. ACF = = 40.2 a 2 = b 2 + c 2 2bc cos A x 2 = (20) 2 + (18) 2 2(20)(18) cos 40.2 = = => x = = 13.2cm The Dublin School of Grinds Page 59

62 Question 2 i) BKA = 60 (360 0 divided into 6 equal parts) ii) Area of triangle KAB = 1 ab sin C 2 = 1 2 (30)(30)sin60 iii) iv) = 389.7km 2 Area of sector KAB = πr 2 ( θ 360 ) = π(30) 2 ( ) = 471.2km 2 Area of triangle KAB % = Area of sector KAB % = 82.7% a 2 = b 2 + c 2 2bc cos A x 2 = (30) 2 + (30) 2 2(30)(30) cos 60 = = 900 => x = 900 = 30m v) All of the sides are 30km so it is an equilateral triangle Question 3 i) In a scalene triangle all the sides and all the angles are different sizes. ii) Area = 1 2 (8)(9)sin60 iii) = 31.18m 2 a 2 = b 2 + c 2 2bc cos A x 2 = (8) 2 + (9) 2 2(8)(9) cos 60 = = 73 => x = 73 = 8.54m The Dublin School of Grinds Page 60

63 Question 4 We can use alternate angles to find two of the angles in the triangle ABG. a) Distance from G to A: Distance from G to B: b) Remember: Time = Distance Speed a sina = b sinb x sin40 = 72 sin73 x = 72sin40 sin73 => x = 48.4m a sina = b sinb x sin67 = 72 sin73 x = 72sin67 sin73 => x = 69.3m Time from G to A: Time from G to B: Time = = 53.8s Time = = 21.7s c) She might not choose the faster route because it will take her further downstream. The Dublin School of Grinds Page 61

64 Question 5 a) a sina = b sinb x sin50 = 8 sin30 x = 8sin50 sin30 => x = 12 b) i) But sinθ = Opposite Hypotenuse = 6 x sinθ = 3 5 So: ii) iii) c) i) 3 5 = 6 x => x = 10 => AC = = x = 36 + x 2 64 = x 2 64 = x 8 = x => AB = 8 sinθ = 3 and cosθ = = 4 5 sin 2 θ +cos 2 θ = ( ) + ( ) = = = 1 a sina = b sinb 35 sinx = 62 sin82 35sin82 = 62sinX => 35sin82 62 = sinx sin 1 ( 35sin82 ) = X 62 => 34 = X => QRS = 34 ii) a 2 = b 2 + c 2 2bc cos A 35 2 = (3x) 2 + (8x) 2 2(3x)(8x) cos = 9x x 2 24x = 49x 2 25 = x 2 => x = 5 The Dublin School of Grinds Page 62

65 Question 6 a) cos60 = 35 x => x = 35 cos60 = 70 => AB = 70cm b) 70 2 = x = x = x = x => x = 60.6cm c) Area of yellow = 1 2 (70)(140)sin60 = 4, Area of blue = 1 2 (70)(70)sin120 = 2, Total Area = 4, , = Area of yellow Fraction of yellow = Total area = 4, = 2 3 The Dublin School of Grinds Page 63

66 Question 7 a) Area = 1 ab sin C 2 = 1 2 (6)(5)sin135 b) i) = 11cm = x = x 2 64 = x 2 ii) iii) c) i) ii) 64 = x => x = 8 sinθ = cosθ = 8 17 cos50 = 17.4 x => x = 17.4 cos50 = 27m => PR = 27m a 2 = b 2 + c 2 2bc cos A x 2 = (15) 2 + (27) 2 2(15)(27) cos 130 = = 1, => x = 1, = 38m => PS = 38m The Dublin School of Grinds Page 64

67 Question 8 i) Remember: Distance = Speed Time ii) 25 minutes = hours Distance = = 160km => bc = 160km a sina = b sinb x sin36 = 160 sin100 x = 160sin36 sin100 => x = 95km => ac = 95km Question 9 a) (i) Angle between blades = 120 (ii)1 blade is 65 metres long. => the radius of the disc is 65m Area = πr 2 = π(65) 2 = 13,273m 2 (iii) From the diagram above we can find the area of the triangle ABC, then multiply this by three to find the area formed by the tips of the blades. Area of a triangle = 1 ab sin C 2 Area of ABC = 1 (65)(65) sin Total Area = 3 ( 1 (65)(65) sin 120 ) 2 = 5488m 2 (iv) First, turn 25 years into minutes: 25 years = ( )minutes = 13,140,000 Next, it only operates 31% of this time: 13,140,000 31% = 13,140, = 4,073,400 Now, it rotates 15 times a minute. So find the total number of revolutions: 4,073, = 61,101,000 = b) Height of the tower = h tan 60 = The Dublin School of Grinds Page 65 h 100 h = 100 tan 60 = 173m

68 c) (i) (ii) tan B = B = tan 1 ( ) = 57 Percentage error = Error True value 100% Error = True value Estimate value = = 3 = 3 Percentage error = Error True value 100 = % = 5.3% Question 10 a) (i) Let x = ED. b) (i) (ii) Let θ = AEB. x 2 = (10) 2 + (1.95) 2 x = (10) 2 + (1.95) 2 x = x = 10.2m => ED = 10.2m cos θ = θ = cos 1 ( ) θ = 31 => AEB = 31 DEB = 180 CED AEB = = 138 (ii) Since it is not a right angled triangle we cannot use Pythagoras. In this case we use the Cosine Rule: a 2 = b 2 + c 2 2bc cos A b = 14m c = 10.2m A = 138 a 2 = (14) 2 + (10.2) 2 2(14)(10.2) cos 138 a 2 = a 2 = a = a = 22.6m The Dublin School of Grinds Page 66

69 Question 11 (a) (i) (ii) (b) Cosine Rule: 16 sin 110 = BC sin 42 16(sin 42 ) => = BC sin 110 => BC = 11.39m Area of a triangle = 1 ab sin C 2 = 1 (16)(11.39) sin 28 2 = a 2 = b 2 + c 2 2bc cos A AB 2 = (10)(16) cos 75 AB 2 = AB = AB = 16.53m The Dublin School of Grinds Page 67

70 Co-ordinate Geometry of the Line Co-ordinate Geometry of the Line is worth between 4% and 6% of the Leaving Certificate. It appears on Paper Finding the distance between two points (using the distance formula) The distance between two points formula is on page 18 of your formula booklet. Distance between two points (x 2 x 1 ) 2 + (y 2 y 1 ) 2 Note: the distance between the points A and B is the same as the length of the line segment that joins them (the line segment [AB]). The distance between the points A and B is represented as AB. Example 1 A(2, 4) and B(1,3) are two points. Find AB. Solution: You are asked for the distance between A and B. Use the formula (x 2 x 1 ) 2 + (y 2 y 1 ) 2. y 2 ) x 1 y 1 x 2 A(2, 4 ) and B(1,3 Write x1, y1, x2, y2 above the numbers before putting them into the formula! Now replace the letters x1, y1, x2 and y2 in the formula with the numbers. Put the numbers in brackets to avoid sign errors. AB = ((1) (2)) 2 + ((3) ( 4)) 2 Put this into your calculator exactly as it appears here. AB = 50 which your calculator will simplify to 5 2 this is your answer! You can leave your answer in surd (square root) form like this. Question 1.1 C( 2,1) and D(1, 3) are two points. Find CD. The Dublin School of Grinds Page 68

71 Question 1.2 P( 2, 3) and Q(0,4) are two points. Find PQ. Question 1.3 The diagram shows a triangle FGH. (i) Write down the co-ordinates (i) of the points F, G and H. G F H (ii) Investigate if the triangle FGH is isosceles. (Remember an isosceles triangle has two sides that are the same length.) The Dublin School of Grinds Page 69

72 Question 1.4 The diagram shows two points A and B. B A C(11,2) and D(6, 2) are two other points. (i) (ii) Plot the points C and D on the diagram and draw the quadrilateral ABCD. Show that ABCD is a parallelogram. The Dublin School of Grinds Page 70

73 2. Finding the midpoint of a line segment (using the midpoint formula) The midpoint of a line segment formula is on page 18 of your formulae and tables booklet. Midpoint of a line segment ( x 1+x 2 2, y 1 +y 2) 2 Example 1 P(4,2) and Q(10,0) are two points. Find the midpoint of [PQ]. Solution: You are asked for the midpoint of a line segment. Use the formula ( x 1+x 2 2, y 1 +y 2). 2 x 1 y 1 x 2 y 2 P(4,2 ) and Q(10,0 ) Now replace the letters x1, y1, x2 and y2 in the formula with the numbers. Again put the numbers in brackets to avoid sign errors. Midpoint is ( (4)+(10) 2, (2)+(0) ) = ( 14, 2 ) = (7,1) this is your answer! Question 2.1 P( 1, 5) and Q(3,1) are two points. Find the midpoint of [PQ]. Question 2.2 A(2, 1) and B( 1,4) are two points. Find the midpoint of [AB]. The Dublin School of Grinds Page 71

74 3. Finding the endpoint of a line segment when given the midpoint If you are given one end point of a line segment and told what the midpoint is, you can find the other end point by using a translation. Example 1 P( 3,2) is one end point of the line segment [PQ]. The midpoint of [PQ] is R( 1,5). Find Q, the other end point of [PQ]. Solution: Work out what happens to the x and y co-ordinates of P when you move to R. Note: this is called the translation PR. The x co-ordinate goes from 3 to 1 so the x co-ordinate increases by 2. The y co-ordinate goes from 2 to 5 so the y co-ordinate increases by 3. So you can now write the translation PR in simple terms. Add 2 to the x co-ordinate, add 3 to the y co-ordinate. If you repeat this translation starting from R, you will get to the other end, Q! add 2 to x add 3 to y ( 3,2) P add 2 to x add 3 to y R( 1,5) Q Adding 2 to the x co-ordinate of R gives you = 1. Adding 3 to the y co-ordinate of R gives you = 8. So the co-ordinates of Q are (1,8) this is your answer! Question 3.1 C(2, 5) is the midpoint of the line segment [AB]. If A is (4,1), find the co-ordinates of B. The Dublin School of Grinds Page 72

75 Question 3.2 The point (1 1 2, 2) is exactly half way between the points (5, 5) and (x,y). Find the values of x and y. The Dublin School of Grinds Page 73

76 4. Finding the slope of a line The slope of a line is a number that tells you two things! The sign tells you whether the line slopes up (goes uphill ) or slopes down (goes downhill ) when reading from left to right. o If the slope is positive, the line goes uphill. o If the slope is negative, the line goes downhill. The number tells you how steep the line is. The larger the number, the steeper the line. line a, slope = 3 line c, slope = 1 3 line b, slope = 1 4 line d, slope = 2 Lines a and b have positive slopes. That means both go uphill. The number for line a is 3. The number for line b is 1 4. This means line a is steeper than line b. Lines c and d have negative slopes. That means both go downhill. The number for line d is 2. The number for line c is 1 3. This means line d is steeper than line c. There are three ways of working out the slope of a line. 1 By using the slope of a line formula. 2 By picking it out of the equation of the line. 3 By reading it from a graph. Using the slope of a line formula The slope of a line formula is on page 18 of your formulae and tables booklet. Slope of a line y 2 y 1 x 2 x 1 To find the slope of a line using this formula, you need two points on the line. You put the x and y co-ordinates of the two points into the formula as before. The slope is represented by the letter m. The Dublin School of Grinds Page 74

77 Example 1 R( 1,1) and S(1, 3) are two points. Find the slope of the line RS. Solution: You are asked for the slope of a line. Use the formula y 2 y 1 x 2 x 1. x 1 y 1 x 2 R( 1,1 ) and S(1, 3 y 2 ) Now replace the letters x1, y1, x2 and y2 in the formula with the numbers. Again put the numbers in brackets to avoid sign errors. Slope of RS = ( 3) (1) (1) ( 1). Put this into your calculator exactly as it appears here. Slope of RS = 2 this is your answer! Question 4.1 P( 2,7) and Q(1,4) are two points. Find the slope of the line PQ. Question 4.2 A(2, 2) and B(8, 1) are two points. Find the slope of the line AB. What does this slope value tell you about the line AB? Question 4.3 P(2,5) and Q(4,k) are two points. The slope of the line PQ is 4. Find the value of k. The Dublin School of Grinds Page 75

78 Picking slope values out of the equation of the line If the equation is written in the form ax + by + c = 0, then the slope of the line is a b. Example 2 Find the slope of the line 2x 3y + 1 = 0. Solution: The equation is in the form ax + by + c = 0. a = the number in front of the x = 2. b = the number in front of the y = 3. The slope is 2 = 2 this is your answer! 3 3 a must be the number in front of the x and b must be the number in front of the y! If the equation is written in the form y = mx + c, then the slope of the line is m (the number in front of the x). This formula is on page 18 of your formulae and tables booklet! Example 3 Find the slope of the line y = 2x 5. Solution: The slope of the line is the number in front of the x = 2 this is your answer! Sometimes, you may need to rearrange the equation of a line to get it into one of these two forms (ax + by + c = 0 or y = mx + c) if you need to find the slope. Example 4 The line k is 3x = 4y 1. Find the slope of k. Solution: You can move everything on the right hand side over onto the left. 3x = 4y 1 becomes 3x 4y + 1 = 0 this is now in the form ax + by + c = 0 a = the number in front of the x = 3. b = the number in front of the y = 4. so the slope is 3 = 3 this is your answer! 4 4 Example 5 The line k is 5y = 6x + 3. Find the slope of k. Solution: You can divide across by 5 to leave a single y on the left hand side. 5y = 6x + 3 becomes y = 6 x + 3 this is now in the form y = mx + c 5 5 The slope of the line is the number in front of the x = 6 this is your answer! 5 Question 4.4 Find the slope of the line 6x + 3y 5 = 0. The Dublin School of Grinds Page 76

79 Question 4.5 The line n is 3x = 5 2y. Find the slope of n. Reading slope values from a graph The slope of a line is given by vertical change horizontal change or rise run. Note: rise can be negative if the line goes downhill. This will give you the expected negative slope value. To find a slope value from a graph, 1 draw in two lines to form a triangle like the one above anywhere on the line, 2 count the horizontal change (the run, which will always be positive), 3 count the vertical change (the rise, which may be positive or negative), 4 put rise over run to get the slope. horizontal change run vertical change rise The Dublin School of Grinds Page 77

80 Example 6 The diagram below shows two lines, m and n. Use the graph to find the slope of m and the slope of n. m n Solution: horizontal change 4 boxes so run = 4 m vertical change goes down by 3 boxes so rise = 3 vertical change goes up by 2 boxes so rise = 2 horizontal change 3 boxes so run = 3 n Slope of m = rise = 2 this is your answer! run 3 Note the positive slope value as the line m is uphill. Slope of n = rise = 3 = 3 this is your answer! run 4 4 Notice the negative slope value as the line n is downhill. The Dublin School of Grinds Page 78

81 Question 4.6 Find the slopes of the lines p, q and r in the diagram below. p q r Question 4.7 The diagram below shows four lines m, n, p, q. m n p q The table below gives the slopes of m, n, p and q. Match each slope to the correct line. Slope Line The Dublin School of Grinds Page 79

82 Question 4.8 The diagram below shows four lines a, b, c and d. a b c d The table below gives the slopes of a, b, c and d. Match each slope to the correct line. Slope Line You can t find the slope values this time because there are no gridlines! 4 3 The Dublin School of Grinds Page 80

83 5. Finding slopes of parallel and perpendicular lines If two lines are parallel, then they have the same slope. To show that two lines are parallel 1 find the slope of each line, 2 show that the two slopes are the same. If the slopes are not the same, the lines are not parallel! If you are told that one line is parallel to another, to find the slope of the second line 1 find the slope of the first line, 2 use this as the slope of the second line. If two lines are perpendicular (meaning that they form a right angle with one another), then the product of their slopes is 1. In simple terms, if you multiply the slopes of two perpendicular lines together, you always get 1 as your answer. To show that two lines are perpendicular 1 find the slope of each line, 2 multiply the two slopes together, 3 show that you get an answer of 1. If you don t get 1 as an answer, then the lines are not perpendicular. If you are told that one line is perpendicular to another, to find the slope of the second line 1 find the slope of the first line, 2 turn this slope upside-down, 3 change the sign of this slope. Example 1 Show that the lines 2x 5y + 3 = 0 and 10y = 2 + 4x are parallel. Solution: Find the slope of each line. 2x 5y + 3 = 0 this is in the form ax + by + c = 0 a = the number in front of the x = 2. b = the number in front of the y = 5. So the slope is a b = 2 5 = y = 2 + 4x You can divide across by 10 to leave a single y on the left hand side. 10y = 2 + 4x becomes y = x this is now in the form y = mx + c So the slope of the line is the number in front of the x = 4 = Both lines have the same slope so they are parallel. Example 2 m is the line 4x + 3y 1 = 0. The line n is perpendicular to m. Find the slope of n. Solution: First find the slope of m. 4x + 3y 1 = 0 this is in the form ax + by + c = 0 a = the number in front of the x = 4. b = the number in front of the y = 3. So the slope of m is a b = 4 3. To find the slope of n, turn this upside-down and change the sign. 4 becomes 3 and change the sign of the slope from negative to positive. 3 4 So the slope of n is 3 this is your answer! 4 The Dublin School of Grinds Page 81

84 Question 5.1 A( 1, 2), B(3,0), C(2,3) and D( 2,1) are four points. Show that AB CD. remember the symbol means is parallel to Question 5.2 P(4,3), Q( 2,0), R( 1,1) and S( 2,3) are four points. Show that PQ RS. remember the symbol means is perpendicular to Question 5.3 p is the line 4x 5y + 3 = 0. q is a line that passes through the points R(2,4) and S(7,8). Is p q? The Dublin School of Grinds Page 82

85 Question 5.4 A( 6,2), B( 2,1) and C(0,9) are the three vertices of a triangle ABC. Investigate whether or not the triangle ABC is right-angled. The Dublin School of Grinds Page 83

86 6. Finding the equation of a line (using the equation formula) The equation of a line formula is on page 18 of your formulae and tables booklet. Equation of a line: y y 1 = m(x x 1 ) To find the equation of a line using this formula, you need any one point on the line and the slope of the line (m). You put the x and y co-ordinates of the point and the value of the slope into the formula as before. Example 1 The line p passes through the point (1, 3) and has a slope of 2 5. Find the equation of the line p. Solution: You are asked for the equation of a line. Use the formula y y 1 = m(x x 1 ). x 1 y 1 (1, 3 ) and m = 2 5. Now replace the letters x1, y1 and m in the formula with the numbers. Again put the numbers in brackets to avoid sign errors. Equation of p: y ( 3) = 2 (x (1)). 5 You need to deal with the fraction on the right hand side first. Multiply both sides by what s below the line in the fraction, in this case 5. When you multiply the right hand side by 5, you get rid of the 5 below the line. You have to multiply the left hand side by 5 too to keep the equation balanced. So you end up with 5(y ( 3)) = 2(x (1)) 5(y + 3) = 2(x 1) now multiply out the brackets 5y + 15 = 2x 2 now move everything onto the same side 5y x + 2 = 0 2x + 5y + 17 = 0 this answer is fine! or you can multiply across by 1 to change all the signs and make the x positive. 2x 5y 17 = 0 this is also fine as your answer! Question 6.1 The line k has a slope of 2 and passes through (3, 5). Find the equation of the line k. It is usual to give the equation of a line with all the terms on the same side of the = sign. The Dublin School of Grinds Page 84

87 Question 6.2 Find the equation of the line that passes through the points (3, 4) and (4, 1). Question 6.3 Find the equation of the line that passes through the points (2, 5) and (7, 8). The Dublin School of Grinds Page 85

88 Question 6.4 Find the equation of a line parallel to x 2y + 1 = 0 that passes through the origin. Question 6.5 r is the line 2x 4y + 3 = 0. s is a line that is parallel to r and passes through the point (1, 1). Find the equation of s. Question 6.6 P( 2,6) and Q(4, 4) are two points. k is a line that passes through the midpoint of [PQ] and is perpendicular to PQ. Find the equation of k. The Dublin School of Grinds Page 86

89 Question 6.7 s is the line 5x y + 17 = 0. t is a line that is perpendicular to s and passes through the point (2,2). Find the equation of t. The Dublin School of Grinds Page 87

90 7. Checking if a point is on a line To check whether or not a point is on a given line, put the x and y co-ordinates of the point into the equation of the line. If these x and y values make the statement true, the point is on the line. If they do not make the statement true, the point is not on the line. Example 1 Are the points A( 2,1) or B(1,5) on the line 2x + y + 3 = 0? Solution: Put the x and y co-ordinates of the point A into the equation of the line. 2( 2) + (1) + 3 = = 0 0 = 0 this is true so the point A is on the line! Put the x and y co-ordinates of the point B into the equation of the line. 2(1) + (5) + 3 = = 0 10 = 0 this is not true so the point B is not on the line! Question 7.1 m is the line 2x 3y 3 = 0. (i) Is the point ( 6, 5) on the line m? (ii) The point (k,1) is on the line m. Find the value of k. The Dublin School of Grinds Page 88

91 Question 7.2 Does the line 2x + y 2 = 0 pass through the origin? Question 7.3 The line n passes through the points (2,3) and (4,4). (i) Find the equation of the line n. (ii) Investigate whether the point ( 2,1) is on the line n. The Dublin School of Grinds Page 89

92 Question 7.4 Investigate whether the points (2,0), ( 1,3) and (0, 4) are collinear. Remember if three points are collinear, it means they are all in a straight line. Question 7.5 The point (2, 1) is on the line 3x + ky 8 = 0. Find the value of k. The Dublin School of Grinds Page 90

93 8. Plotting lines on a co-ordinate diagram In order to plot a line, you need to find two points on the line, plot those points on a co-ordinate diagram and then join them together with a ruler. Usually, the easiest way to find two points is to find where the line cuts the axes. To find where a line cuts the x-axis, put y = 0 and find the x co-ordinate. The x co-ordinate of this point is called the x-intercept. To find where a line cuts the y-axis, put x = 0 and find the y co-ordinate. The y co-ordinate of this point is called the y-intercept. Note: when the equation of a line is written in the form y = mx + c, you know that m (the number in front of the x) is the slope of the line. The value of c (the number by itself) is the y-intercept (where the line cuts the y-axis). If m is the line x + 2y 4 = 0, then 2y = x + 4 then y = 1 x + 2 this is now in the form y = mx + c 2 The slope is the number in front of the x = 1. 2 But notice that the number on its own (which is 2) is where the line cuts the y-axis! Example 1 On a co-ordinate diagram, plot the line 3x 5y 15 = 0. Solution: First find where the line cuts the x-axis by putting y = 0. 3x 5(0) 15 = 0 3x 0 15 = 0 3x 15 = 0 3x = 15 x = 5 So the point is (5,0) remember you put the y co-ordinate = 0 to find this point Next find where the line cuts the y-axis by putting x = 0. 3(0) 5y 15 = 0 0 5y 15 = 0 5y 15 = 0 5y = 15 x = 3 So the point is (0, 3) remember you put the x co-ordinate = 0 to find this point Now plot these two points on a co-ordinate diagram. Join them together with a ruler and that is the line you were asked to plot. The Dublin School of Grinds Page 91

94 Question 8.1 On a co-ordinate diagram, plot the line 2x + 5y 10 = 0. Question 8.2 On a co-ordinate diagram, plot the line 3x 5y + 12 = 0. The Dublin School of Grinds Page 92

95 Any line parallel to the x-axis will have an equation which will be y = (some number). The diagram shows the lines y = 3 and y = 4. Note: lines parallel to the x-axis are horizontal and horizontal lines are flat. They have no slope so their slope value = 0. Any line parallel to the y-axis will have an equation which will be x = (some number). The diagram shows the lines x = 1 and x = 2. Note: lines parallel to the x-axis are vertical. Their slopes are undefined. x = 2 x = 1 y = 3 The equation of the x-axis is y = 0. The equation of the y-axis is x = 0. y = 4 The Dublin School of Grinds Page 93

96 9. Finding the point of intersection of two lines Unless they are parallel, two lines will intersect at exactly one point. There are two different ways of finding the point of intersection of two lines. 1 By plotting the lines on a co-ordinate diagram and reading the point of intersection off the graph you have drawn. 2 By solving the simultaneous equations of the two lines. Example 1 Find the point of intersection of the following lines: x + y 2 = 0 and 2x 3y + 16 = 0 Solution: First put the numbers on the right hand side in each equation. x + y = 2 2x 3y = 16 You need to get to a point where either the x s or the y s in these equations are equal in number but opposite in sign so that when you add them together, one of the letters will cancel out. You can multiply everything on both sides of the top equation by 3. So x + y = 2 becomes 3x + 3y = 6 Now there is a 3y in both equations but it is positive in one, negative in the other. So now add the two equations. 3x + 3y = 6 2x 3y = 16 5x 3y = 10 x = 2 Put this x value back into either of the original equations to find the value for y. ( 2) + y = y = 2 y = 4 So the point of intersection is ( 2,4) this is your answer! Question 9.1 Find the point of intersection of the lines 2x + 3y 7 = 0 and 5x 2y 8 = 0. The Dublin School of Grinds Page 94

97 Question 9.2 p is the line x + 2y 4 = 0 and q is the line 4x 5y 29 = 0. Find p q. remember the symbol means intersection Question 9.3 m is the line 3x + y 6 = 0 and n is the line x 2y + 5 = 0. (i) On the same axes and scales, plot the lines m and n. The Dublin School of Grinds Page 95

98 (ii) From your graph, identify the point of intersection of m and n. (iii) Confirm your answer by solving the simultaneous equations. The Dublin School of Grinds Page 96

99 10. Finding the area of a triangle using the formula If you have drawn a graph and you are asked to find the area of a triangle, check if one of the sides is either on or parallel to either the x-axis or y-axis. If so, take this side to be the base of the triangle, you should be able to read the length of this base from your graph, you should be able to read the height of the triangle too from your graph, use the formula for the area of a triangle (area = 1 base height). 2 Example 1 The diagram shows the triangle XYZ. Find the area of XYZ. X Solution: Notice the side [XY] is parallel to the y-axis. You can call this the base and you can see that it is 5 units long (count the boxes!). Notice the point Z is 4 units away from this base. So the height is 4. Z So the area of XYZ = 1 base height 2 So the area of XYZ = 1 2 (5)(4) = 10 units2. Y The Dublin School of Grinds Page 97

100 Question 10.1 The diagram below shows two lines p and q. Find the area of the triangle formed by p, q and the x-axis. p q If you have not drawn a diagram (or been given one) or if none of the sides of the triangle are parallel to one of the axes, you will have to use the area of a triangle formula. The area of a triangle formula is on page 18 of your formulae and tables booklet. Area of a triangle 1 2 x 1y 2 x 2 y 1 This formula only works when one of the vertices (corners) of the triangle is (0,0)! Put the x and y co-ordinates of the other two points (not (0,0)) into the formula. If none of the vertices are (0,0), you must move one of the corners to (0,0) using a translation. You must then remember to move the other two corners by the same translation and then use the formula. Remember the straight lines ( ) simply mean that if you get a negative answer, ignore the minus sign!...because a shape can t have a negative area The Dublin School of Grinds Page 98

101 Example 2 P(2,5), Q(4, 2) and R( 1,3) are three points. Find the area of the triangle PQR. Solution: None of the corners are (0,0). Pick any one of the corners and move it to (0,0). Work out the translation that moves it there like we did in section 3 above. "subtract 2 from x, subtract 5 from y" P(2,5) (0,0) Now remember to move the other two corners by the same translation! "subtract 2 from x, subtract 5 from y" Q(4, 2) (2, 7) "subtract 2 from x, subtract 5 from y" R( 1,3) ( 3, 2) Now use the formula 1 x 2 1y 2 x 2 y 1. y 2 ) x 1 y 1 x 2 (2, 7 ) and ( 3, 2 Now replace the letters x1, y1, x2 and y2 in the formula with the numbers. Put the numbers in brackets to avoid sign errors. Area of PQR = 1 (2)( 2) ( 3)( 7) 2 Area of PQR = = 1 2 (25) = 25 2 or units2 this is your answer! Notice we dropped the minus sign from the 25 because that is what the straight lines ( ) mean. This gives us a positive answer. Question 10.2 A( 5,3), B(4, 6) and C(0,0) are the three vertices of a triangle ABC. Find the area of the triangle ABC. The Dublin School of Grinds Page 99

102 Question 10.3 P(6,5), Q(3, 3) and R(4,0) are the three vertices of a triangle PQR. Find the area of the triangle PQR. Question 10.4 D( 1,2), E(2,5) and F(7, 3) are the three vertices of a triangle DEF. Find the area of the triangle DEF. The Dublin School of Grinds Page 100

103 Question 10.5 P( 4,3), Q(1,1), R(5, 1) and O(0,0)are the four vertices of a quadrilateral PQRO. Find the area of PQRO. The Dublin School of Grinds Page 101

104 11. Using translations We have already used translations twice! We used them in section 3 to find the end point of a line segment when given its midpoint. We used them again in section 10 to move the corners of a triangle so one of them was (0,0) so we could use the area of a triangle formula. Translations can also be used to find the missing corner of a square, rectangle or parallelogram. You must be careful to maintain what is called the cyclic order of points. If a parallelogram is named ABCD, you must encounter the points A, B, C and D in that order as you move around the perimeter of the shape! The diagram shows a triangle ABC. If you want to add a fourth point D to form a parallelogram, there are three different places you could put it, as you see! A 2 3 B C 1 If you put D in position 1, the name of the parallelogram would be ABDC. If you put D in position 2, the name of the parallelogram would be ADBC. If you put D in position 3, the name of the parallelogram would be ABCD. So the name of the parallelogram in the question will tell you where the fourth corner should go. You can then find the exact location of this point using translations as before. The Dublin School of Grinds Page 102

105 Example 1 P( 1,5), Q(2, 3) and R(9, 2) are three points. If PQRS is a parallelogram, find the co-ordinates of S. Solution: P S First draw a rough sketch to see where the point would have to go to give a parallelogram named PQRS. Because the sides are parallel and the same length, moving from Q to P is the same as moving from R to S! So moving R by the translation QP will give you the coordinates of S. Work out what happens to the x and y co-ordinates of Q when you move to P. Note: this is the translation QP. Q R The x co-ordinate goes from 2 to 1 so the x co-ordinate decreases by 3. The y co-ordinate goes from 3 to 5 so the y co-ordinate increases by 8. So QP is subtract 3 from the x co-ordinate, add 8 to the y co-ordinate. If you repeat this translation again starting at R, you will get to S! Subtract 3 from the x co-ordinate of R gives you 9 3 = 6. Add 8 to the y co-ordinate of R gives you = 6. So the co-ordinates of S are (6,6) this is your answer! Question 11.1 A( 2,3), B(5,7) and C(6, 1) are three points. If ABXC is a parallelogram, find the co-ordinates of X. The Dublin School of Grinds Page 103

106 12. Past and probable exam questions Question 1 A(6, 1), B(12, 3), C(8,5) and D(2,7) are four points. (a) Plot the four points on the diagram below. (b) Describe two different ways of showing, using co-ordinate geometry techniques, that the points form a parallelogram ABCD. First Method: Second Method: The Dublin School of Grinds Page 104

107 (c) Use one of the ways you have described to show that ABCD is a parallelogram. The Dublin School of Grinds Page 105

108 Question 2 A( 4,1) and B(2, 3) are two points. (i) Calculate AB. (ii) Find C, the midpoint of [AB]. (iii) Find the slope of the line AB. (iv) Is the line 2x = 4 3y parallel to AB? Explain your answer. The Dublin School of Grinds Page 106

109 (v) Find the equation of the line AB. (vi) The line k is perpendicular to AB and passes through C. Find the equation of k. (vii) D is the point (3,5). Show that D k. remember the symbol means is an element of This is another way of asking you to show the point D is on the line k. The Dublin School of Grinds Page 107

110 (viii) Find the area of the triangle ABD. The Dublin School of Grinds Page 108

111 Question 3 (a) l is the line 3x + 2y + 18 = 0. Find the slope of l. (b) The line k is perpendicular to l and cuts the x-axis at the point (7,0). Find the equation of k. (c) Find the co-ordinates of the point of intersection of the lines l and k. The Dublin School of Grinds Page 109

112 Question 4 The diagram below shows the lines p, q, r and s. q s p r The table below gives the slopes of p, q, r and s. Match each slope to the correct line. Slope Line The Dublin School of Grinds Page 110

113 Question 5 The diagram below shows a line k. k (i) From the graph, find the slope of k. (ii) Find the equation of k. The Dublin School of Grinds Page 111

114 (iii) m is the line 2x + 3y 12 = 0. Plot the line m on the same diagram above. (iv) From your graph, find the point of intersection of k and m. (v) Confirm your answer by solving the simultaneous equations. The Dublin School of Grinds Page 112

115 Question 6 The points A(6,1)and B(2, 1) are shown on the diagram. (i) Find the equation of the line AB. A B (b) The line AB crosses the y-axis at C. Find the co-ordinates of C. (c) Find the ratio AB, giving your answer in the form p, where p and q are whole numbers. AC q The Dublin School of Grinds Page 113

116 Question 7 The Dublin School of Grinds Page 114

117 The Dublin School of Grinds Page 115

118 Question 8 The Dublin School of Grinds Page 116

119 13. Solutions to Co-ordinate geometry of the line Question 1.1 x 1 y 1 x 2 C( 2,1 ) and D(1, 3 y 2 ) CD = ((1) ( 2)) 2 + (( 3) (1)) 2 CD = 25 = 5 Question 1.2 x 1 y 1 x 2 P( 2, 3 ) and Q(0,4 y 2 ) PQ = ((0) ( 2)) 2 + ((4) ( 3)) 2 PQ = 53 you can leave your answer in surd (square root) form like this Question 1.3 (i) F is ( 3,2), G is (0,7) and H is (2, 1). (ii) Find the lengths of the three sides and see if two are the same. First find FG. x 1 y 1 x 2 F( 3,2 ) and G(0,7 y 2 ) FG = ((0) ( 3)) 2 + ((7) (2)) 2 FG = 34 Next find GH. x 1 y 1 x 2 G(0,7 ) and H(2, 1 y 2 ) GH = ((2) (0)) 2 + (( 1) (7)) 2 GH = 68 Finally find FH. x 1 y 1 x 2 F( 3,2 ) and H(2, 1 y 2 ) FH = ((2) ( 3)) 2 + (( 1) (2)) 2 FH = 34 Notice that two sides, [FG] and [FH], are the same length. So FGH is isosceles. The Dublin School of Grinds Page 117

120 Question 1.4 (i) B A C (ii) Find the lengths of the four sides and show that opposite pairs of sides are the same length. First find AB. D y 2 ) x 1 y 1 x 2 A( 3,1 ) and B(2,5 AB = ((2) ( 3)) 2 + ((5) (1)) 2 AB = 41 Now find CD the length of the side opposite [AB] x 1 y 1 x 2 C(11,2 ) and D(6, 2 y 2 ) CD = ((6) (11)) 2 + (( 2) (2)) 2 CD = 41 So one pair of opposite sides are the same length. Now find AD. x 1 y 1 x 2 A( 3,1 ) and D(6, 2 y 2 ) AD = ((6) ( 3)) 2 + (( 2) (1)) 2 AD = 90 which your calculator will simplify to 3 10 Now find BC the length of the side opposite [AD] x 1 y 1 x 2 y 2 B(2,5 ) and C(11,2 ) BC = ((11) (2)) 2 + ((2) (5)) 2 BC = 90 which again your calculator will simplify to 3 10 So the other pair of opposite sides are the same length. As opposite sides are the same length, ABCD is a parallelogram. The Dublin School of Grinds Page 118

121 Question 2.1 x 1 y 1 x 2 P( 1, 5 ) and Q(3,1 Midpoint is ( ( 1)+(3) 2 y 2 ), ( 5)+(1) 2 ) = ( 2, 4 ) = (1, 2) 2 2 Question 2.2 x 1 y 1 x 2 A(2, 1 ) and B( 1,4 Midpoint is ( (2)+( 1) 2 y 2 ), ( 1)+(4) ) = ( 1, 3 ) this is your answer! You can leave your answer like this as it will not simplify any further. Question 3.1 Work out the translation AC. You are moving from A(4,1) to C(2, 5). The x co-ordinate goes from 4 to 2 so the x co-ordinate decreases by 2. The y co-ordinate goes from 1 to 5 so the y co-ordinate decreases by 6. So AC is subtract 2 from the x co-ordinate, subtract 6 from the y co-ordinate. Repeat this translation from C. Subtracting 2 from the x co-ordinate of C gives you 2 2 = 0. Subtracting 6 from the y co-ordinate of C gives you 5 6 = 11. So the co-ordinates of B are (0, 11). Question 3.2 Work out the translation that moves a point from (5, 5) to (1 1 2, 2). The x co-ordinate goes from 5 to 1 1 so the x co-ordinate decreases by The y co-ordinate goes from 5 to 2 so the y co-ordinate increases by 3. So the translation is subtract 3 1 from the x co-ordinate, add 3 to the y co-ordinate. 2 Repeat this translation from the midpoint (1 1 2, 2). Subtracting from the x co-ordinate gives you = 2. Adding 3 to the y co-ordinate gives you = 1. So the co-ordinates of the other end point are ( 2,1). Question 4.1 Use the slope of a line formula. x 1 y 1 x 2 P( 2,7 ) and Q(1,4 y 2 ) Slope of PQ = (4) (7) (1) ( 2) = 3 3 = 1 Question 4.2 Use the slope of a line formula. x 1 y 1 x 2 A(2, 2 ) and B(8, 1 y 2 ) Slope of AB = ( 1) ( 2) (8) (2) = 1 6 The line AB goes uphill because the slope is positive. The number 1 is not very big so the line AB is not very steep 6 The Dublin School of Grinds Page 119

122 Question 4.3 Use the slope of a line formula. x 1 y 1 x 2 P(2,5 ) and Q(4,k y 2 ) Slope of PQ = (k) (5) = k 5 (4) (2) 2 we can t work this out any further unless we know k The question tells you the slope should equal 4. So put the slope you have worked out equal to 4 and solve to find k! So k 5 2 So k 5 = 8 So k = 3 = 4 multiply across by 2 Question 4.4 6x + 3y 5 = 0 this is in the form ax + by + c = 0 a = the number in front of the x = 6. b = the number in front of the y = 3. so the slope is 6 3 = 2 Question 4.5 Move everything on the right hand side over onto the left. 3x = 5 2y becomes 3x + 2y 5 = 0 this is now in the form ax + by + c = 0 a = the number in front of the x = 3. b = the number in front of the y = 2. so the slope is 3 2 Question 4.6 p q For p, rise = 4 and run = 3. So the slope of p = rise run = 4 3. For q, rise = 2 and run = 5. r So the slope of q = rise run = 2 5. For r, rise = 1 and run = 4. So the slope of r = rise run = 1 4 = 1 4. The Dublin School of Grinds Page 120

123 Question 4.7 m p n q Slope Line n m 3 p 1 2 q For m, rise = 5 and run = 1. So the slope of m = rise run = 5 1 = 5. For n, rise = 3 and run = 4. So the slope of n = rise run = 3 4. For p, rise = 3 and run = 1. So the slope of p = rise run = 3 1 = 3. For q, rise = 1 and run = 2. So the slope of q = rise run = 1 2 = 1 2. Just match each line with the correct slope value in the table. Question 4.8 Slope 2 3 Line c a and d both go downhill so the two negative slope values must belong to a and d. a is steeper than d so it must have the bigger number. Since 4 3 > 1 5, a must be 4 3 and d must be d b and c both go uphill so the two positive slope values must belong to b and c. b is steeper than c so it must have the bigger number. 5 4 b Since 5 4 > 2 3, b must be 5 4 and c must be a The Dublin School of Grinds Page 121

124 Question 5.1 First find the slope of AB. Use the slope of a line formula. x 1 y 1 x 2 A( 1, 2 ) and B(3,0 y 2 ) Slope of AB = (0) ( 2) (3) ( 1) = 1 2 Next find the slope of CD. x 1 y 1 x 2 C(2,3 ) and D( 2,1 y 2 ) Slope of CD = (1) (3) ( 2) (2) = 1 2 The lines AB and CD have the same slopes so they are parallel. Question 5.2 First find the slope of PQ. Use the slope of a line formula. x 1 y 1 x 2 P(4,3 ) and Q( 2,0 y 2 ) Slope of PQ = (0) (3) ( 2) (4) = 1 2 Next find the slope of RS. x 1 y 1 x 2 R( 1,1 ) and S( 2,3 y 2 ) Slope of RS = (3) (1) = 2 = 2 ( 2) ( 1) 1 Multiplying these two slopes together gives 1 2 = 1 2 Multiplying the slopes of PQ and RS together gives an answer of 1 so the lines are perpendicular. Question 5.3 First find the slope of p. 4x 5y + 3 = 0 this is in the form ax + by + c = 0 a = the number in front of the x = 4. b = the number in front of the y = 5. so the slope of p is 4 5 = 4 5 Next find the slope of q. Use the slope of a line formula. x 1 y 1 x 2 R(2,4 ) and S(7,8 y 2 ) Slope of PQ = (8) (4) (7) (2) = 4 5 The lines p and q have the same slopes so they are parallel. The Dublin School of Grinds Page 122

125 Question 5.4 Find the slopes of the three sides and see if any two sides are perpendicular. If two of the sides are perpendicular, they form a right angle so the triangle would be right-angled. First find the slope of AB. Use the slope of a line formula. x 1 y 1 x 2 A( 6,2 ) and B( 2,1 y 2 ) (1) (2) Slope of AB = = 1 ( 2) ( 6) 4 Next find the slope of BC. x 1 y 1 x 2 B( 2,1 ) and C(0,9 y 2 ) Slope of BC = (9) (1) (0) ( 2) = 4 Multiplying these two slopes together gives = 1 Multiplying the slopes of AB and BC together gives an answer of 1 so these two sides are perpendicular. This means ABC is right-angled. Note: because you have already found two sides that are perpendicular, you don t need to find the slope of the third side. Question 6.1 Use the equation of a line formula. x 1 y 1 (3, 5 ) and m = 2. Equation of k is y ( 5) = 2(x (3)) y + 5 = 2(x 3) now multiply out the brackets y + 5 = 2x + 6 now move everything onto the same side y x 6 = 0 2x + y 1 = 0 Question 6.2 First find the slope of the line. Use the slope of a line formula. x 1 y 1 x 2 (3, 4 ) and (4, 1 Slope = ( 1) ( 4) (4) (3) y 2 ) = 3 Now use the equation of a line formula. x 1 y 1 (3, 4 ) and m = 3. Equation of k is y ( 4) = 3(x (3)) y + 4 = 3(x 3) now multiply out the brackets y + 4 = 3x 9 now move everything onto the same side y + 4 3x + 9 = 0 3x + y + 13 = 0 or 3x y 13 = 0 The Dublin School of Grinds Page 123

126 Question 6.3 First find the slope of the line. Use the slope of a line formula. x 1 y 1 x 2 (2, 5 ) and (7,8 y 2 ) Slope = (8) ( 5) (7) (2) = 13 5 Now use the equation of a line formula. x 1 y 1 (2, 5 ) and m = Equation is y ( 5) = 13 (x (2)) multiply across by 5 5 5(y ( 5)) = 13(x (2)) 5(y + 5) = 13(x 2) now multiply out the brackets 5y + 25 = 13x 26 now move everything onto the same side 5y x + 26 = 0 13x + 5y + 51 = 0 or 13x 5y 51 = 0 Question 6.4 First find the slope of the line. x 2y + 1 = 0 this is in the form ax + by + c = 0 a = the number in front of the x = 1. not zero! b = the number in front of the y = 2. so the slope of p is 1 2 = 1 2 Now use the equation of a line formula. x 1 y 1 (0,0 ) and m = 1. use the same slope value as the lines are parallel 2 Equation is y (0) = 1 (x (0)) multiply across by 2 2 2(y (0)) = 1(x (0)) 2(y) = 1(x) 2y = x now move everything onto the same side 2y x = 0 x + 2y = 0 or x 2y = 0 The Dublin School of Grinds Page 124

127 Question 6.5 First find the slope of r. 2x 4y + 3 = 0 this is in the form ax + by + c = 0 a = the number in front of the x = 2. b = the number in front of the y = 4. so the slope of r is 2 4 = 1 2. Now use the equation of a line formula. x 1 y 1 (1, 1 ) and m = 1 2. Equation of s is y ( 1) = 1 (x (1)) multiply across by 2 2 2(y ( 1)) = 1(x (1)) 2(y + 1) = 1(x 1) now multiply out the brackets 2y + 2 = x 1 now move everything onto the same side 2y + 2 x + 1 = 0 x + 2y + 3 = 0 or x 2y 3 = 0 Question 6.6 First find the midpoint of [PQ]. Use the midpoint formula. x 1 y 1 x 2 P( 2,6 ) and Q(4, 4 Midpoint is ( ( 2)+(4) 2 y 2 ) Next find the slope of PQ. Use the slope of a line formula., (6)+( 4) ) = ( 2, 1 ) = (1,1) x 1 y 1 x 2 P( 2,6 ) and Q(4, 4 y 2 ) Slope = ( 4) (6) (4) ( 2) = 5 3 The line k is perpendicular to PQ. So find the slope of k by turning the slope of PQ upside-down and changing its sign. 5 becomes 3 and change the sign of the slope from negative to positive. 3 5 So the slope of k is 3 5. Now use the equation of a line formula. x 1 y 1 (1,1 ) and m = 3. 5 Equation of s is y (1) = 3 5 (x (1)) multiply across by 5 5(y (1)) = 3(x (1)) 5(y 1) = 3(x 1) now multiply out the brackets 5y 5 = 3x 3 now move everything onto the same side 5y 5 3x + 3 = 0 3x + 5y 2 = 0 or 3x 5y + 2 = 0 The Dublin School of Grinds Page 125

128 Question 6.7 First find the slope of s. 5x y + 17 = 0 this is in the form ax + by + c = 0 a = the number in front of the x = 5. b = the number in front of the y = 1. so the slope of s is 5 1 = 5. The line t is perpendicular to s. So find the slope of t by turning the slope of s upside-down and changing its sign. Remember that 5 is the same as becomes 1 and change the sign of the slope from positive to negative. 1 5 So the slope of t is 1. 5 Now use the equation of a line formula. x 1 y 1 (2,2 ) and m = 1. 5 Equation of t is y (2) = 1 (x (2)) multiply across by 5 5 5(y (2)) = 1(x (2)) 5(y 2) = 1(x 2) now multiply out the brackets 5y 10 = x + 2 now move everything onto the same side 5y 10 + x 2 = 0 x + 5y 12 = 0 Question 7.1 (i) Put the x and y co-ordinates of ( 6, 5) into the equation of m. 2( 6) 3( 5) 3 = = 0 0 = 0 this is true! So the point is on the line m. (ii) If you put the x and y co-ordinates of (k,1) into the equation of m, the statement is true because the point is on the line. 2(k) 3(1) 3 = 0 2k 3 3 = 0 2k 6 = 0 2k = 6 k = 3 Question 7.2 Put the x and y co-ordinates of the origin (0,0) into the equation of the line. 2(0) + (0) 2 = = 0 2 = 0 this is not true! So the origin is not on the line. The Dublin School of Grinds Page 126

129 Question 7.3 (i) First find the slope of n. Use the slope of a line formula. x 1 y 1 x 2 y 2 (2,3 ) and (4,4 ) Slope of n = (4) (3) (4) (2) = 1 2 Now use the equation of a line formula. x 1 y 1 (2,3 ) and m = 1. 2 Equation of n is y (3) = 1 (x (2)) multiply across by 2 2 2(y (3)) = 1(x (2)) 2(y 3) = 1(x 2) now multiply out the brackets 2y 6 = x 2 now move everything onto the same side 2y 6 x + 2 = 0 x + 2y 4 = 0 or x 2y + 4 = 0 (ii) Put the x and y co-ordinates of ( 2,1) into the equation of n. ( 2) 2(1) + 4 = = 0 0 = 0 this is true! So the point is on the line n. Question 7.4 This question is essentially identical to the previous one. Find the equation of the line that goes through two of the three points. Then check if the third point is on that line too. First find the slope of the line that passes through (2,0) and ( 1,3). Use the slope of a line formula. x 1 y 1 x 2 y 2 (2,0 ) and ( 1,3 ) Slope = (3) (0) ( 1) (2) = 1 Now use the equation of a line formula. x 1 y 1 (2,0 ) and m = 1. Equation is y (0) = 1(x (2)) y = 1(x 2) now multiply out the brackets on the right hand side y = x + 2 now move everything onto the same side y + x 2 = 0 x + y 2 = 0 Put the x and y co-ordinates of (0, 4) into this equation. (0) + ( 4) 2 = = 0 6 = 0 this is not true! So the point (0, 4) is not on the same line as the other two. This means the three points are not collinear. The Dublin School of Grinds Page 127

130 Question 7.5 Put the x and y co-ordinates of (2, 1) into the equation. 3(2) + k( 1) 8 = 0 6 k 8 = 0 k = 2 k = 2 Question 8.1 You need to find two points on the line 2x + 5y 10 = 0. First find where the line cuts the x-axis by putting y = 0. 2x + 5(0) 10 = 0 2x 0 10 = 0 2x 10 = 0 2x = 10 x = 5 So the point is (5,0) Next find where the line cuts the y-axis by putting x = 0. 2(0) + 5y 10 = y 10 = 0 5y 10 = 0 5y = 10 y = 2 So the point is (0,2) Now plot these two points and join them together. The Dublin School of Grinds Page 128

131 Question 8.2 You need to find two points on the line 3x 5y + 12 = 0. First find where the line cuts the x-axis by putting y = 0. 3x + 5(0) + 12 = 0 3x = 0 3x + 12 = 0 3x = 12 x = 4 So the point is ( 4,0) Next find where the line cuts the y-axis by putting x = 0. 3(0) 5y + 12 = 0 0 5y + 12 = 0 5y + 12 = 0 5y = 12 y = 12 = 2 2 = So the point is (0,2 4) Now plot these two points and join them together. Question 9.1 2x + 3y 7 = 0 5x 2y 8 = 0 Move the number across onto the right hand side in each equation. 2x + 3y = 7 5x 2y = 8 Multiply everything on both sides of the top equation by 2. Multiply everything on both sides of the bottom equation by 3. 2x + 3y = 7 becomes 4x + 6y = 14 5x 2y = 8 becomes 15x 6y = 24 add the equations together 19x + 6 = 38 x = 2 Now find the y co-ordinate by putting this x value into one of the equations. 2(2) + 3y = y = 7 3y = 3 y = 1 So the point of intersection is (2,1) The Dublin School of Grinds Page 129

132 Question 9.2 x + 2y 4 = 0 4x 5y 29 = 0 Move the number across onto the right hand side in each equation. x + 2y = 4 4x 5y = 29 Multiply everything on both sides of the top equation by 5. Multiply everything on both sides of the bottom equation by 2. x + 2y = 4 becomes 5x + 10y = 20 4x 5y = 29 becomes 8x 10y = 58 add the equations together 13x + 6 = 78 x = 6 Now find the y co-ordinate by putting this x value into one of the equations. (6) + 2y 4 = y 4 = 0 2y = 2 y = 1 So the point of intersection is (6, 1) Question 9.3 (i) You need to find two points on the line 3x + y 6 = 0. First find where the line cuts the x-axis by putting y = 0. 3x + (0) 6 = 0 3x = 0 3x 6 = 0 3x = 6 x = 2 So the point is (2,0) Next find where the line cuts the y-axis by putting x = 0. 3(0) + y 6 = y 6 = 0 y 6 = 0 y = 6 So the point is (0,6) Now plot these two points and join them together. Next you need to find two points on the line x 2y + 5 = 0. First find where the line cuts the x-axis by putting y = 0. x 2(0) + 5 = 0 x = 0 x + 5 = 0 x = 5 So the point is ( 5,0) The Dublin School of Grinds Page 130

133 Next find where the line cuts the y-axis by putting x = 0. (0) 2y + 5 = 0 0 2y + 5 = 0 2y + 5 = 0 2y = 5 y = 5 2 = = 2 5 So the point is (0,2 5) Now plot these two points and join them together. (ii) From the graph, the two lines intersect at the point (1,3). (iii) 3x + y 6 = 0 x 2y + 5 = 0 Move the number across onto the right hand side in each equation. 3x + y = 6 x 2y = 5 Multiply everything on both sides of the top equation by 2. Leave the bottom equation alone. 3x + y = 6 becomes 6x + 2y = 12 4x 5y = 29 becomes x 2y = 5 add the equations together 7x + 6 = 7 x = 1 Now find the y co-ordinate by putting this x value into one of the equations. (1) 2y = 5 1 2y = 5 2y = 6 y = 3 So the point of intersection is (1,3) same answer as part (ii) The Dublin School of Grinds Page 131

134 Question 10.1 One side is on the x-axis. This is the base and is 13 units long. The top point of the triangle is 6 units away from the base so the height is 6. So the area = 1 base height = 1 (13)(6) = 39 units2 2 2 Question 10.2 One of the corners C is (0,0). Put the two points that are not (0,0), A and B, into the area of a triangle formula. x 1 y 1 x 2 A( 5,3 ) and B(4, 6 y 2 ) So the area of ABC = 1 ( 5)( 6) (4)(3) = = 1 18 = 9 units Question 10.3 Use the area of a triangle formula. None of the corners are at (0,0). Move any one corner to (0,0) using a translation. Then move the other two corners with the same translation. "subtract 4 from x, don't change y" R(4,0) "subtract 4 from x, don't change y" (0,0) Q(3, 3) ( 1, 3) "subtract 4 from x, don't change y" P(6,5) (2,5) Now put the two points that are not (0,0) into the formula. y 2 ) x 1 y 1 x 2 ( 1, 3 ) and B(2,5 So the area of PQR = 1 2 ( 1)(5) (2)( 3) = = = 1 2 units2 Question 10.4 Use the area of a triangle formula. None of the corners are at (0,0). Move any one corner to (0,0) using a translation. Then move the other two corners with the same translation. "add 1 to x, subtract 2 from y" D( 1,2) "add 1 to x, subtract 2 from y" (0,0) E(2,5) (3,3) "add 1 to x, subtract 2 from y" F(7, 3) (8, 5) Now put the two points that are not (0,0) into the formula. y 2 ) x 1 y 1 x 2 (3,3 ) and B(8, 5 So the area of DEF = 1 2 (3)( 5) (8)(3) = = = 39 2 units2 The Dublin School of Grinds Page 132

135 Question 10.5 Divide the quadrilateral into two triangles as shown below. P Q O R First find the area of PQO. One of the corners O is (0,0). Put the two points that are not (0,0), P and Q, into the area of a triangle formula. x 1 y 1 x 2 P( 4,3 ) and Q(1,1 y 2 ) So the area of PQO = 1 ( 4)(1) (1)(3) = = 1 7 = units2 First find the area of RQO. One of the corners O is (0,0). Put the two points that are not (0,0), R and Q, into the area of a triangle formula. x 1 y 1 x 2 R(5, 1 ) and Q(1,1 y 2 ) So the area of RQO = 1 (5)(1) (1)( 1) = = 1 6 = 3 units So the total area of PQRO = area PQO + area RQO So the total area of PQRO = = 13 2 or units2 The Dublin School of Grinds Page 133

136 Question 11.1 Draw a rough sketch to show you where the missing point should go. B A X C Moving from A to B is the same as moving from C to X! So moving C by the translation AB will give you the co-ordinates of X. Work out what happens to the x and y co-ordinates of A when you move to B. The x co-ordinate goes from 2 to 5 so the x co-ordinate increases by 7. The y co-ordinate goes from 3 to 7 so the y co-ordinate increases by 4. So AB is add 7 to the x co-ordinate, add 4 to the y co-ordinate. If you repeat this translation again starting at C, you will get to X. Add 7 to the x co-ordinate of C gives you = 13. Add 4 to the y co-ordinate of C gives you = 3. So the co-ordinates of X are (13,3) The Dublin School of Grinds Page 134

137 Past and probable exam questions solutions Question 1 (a) D C A B (b) (c) The simplest two to use are 1 use the distance between two points formula to show that opposite sides are the same length. 2 use the slope of a line formula to show that opposite sides are parallel. Use the distance formula to show that opposite sides are the same length. First find AB. x 1 y 1 x 2 A(6, 1 ) and B(12, 3 y 2 ) AB = ((12) (6)) 2 + (( 3) ( 1)) 2 AB = 40 which your calculator will simplify to 2 10 Now find CD the length of the side opposite [AB] x 1 y 1 x 2 y 2 C(8,5 ) and D(2,7 ) CD = ((2) (8)) 2 + ((7) (5)) 2 CD = 40 which again your calculator will simplify to 2 10 So one pair of opposite sides are the same length. Now find AD. x 1 y 1 x 2 A(6, 1 ) and D(2,7 y 2 ) AD = ((2) (6)) 2 + ((7) ( 1)) 2 AD = 80 which your calculator will simplify to 4 5 Now find BC the length of the side opposite [AD] x 1 y 1 x 2 y 2 B(12, 3 ) and C(8,5 ) BC = ((8) (12)) 2 + ((5) ( 3)) 2 BC = 80 which again your calculator will simplify to 4 5 So the other pair of opposite sides are the same length. As opposite sides are the same length, ABCD is a parallelogram. The Dublin School of Grinds Page 135

138 Question 2 (i) Use the distance formula. x 1 y 1 x 2 A( 4,1 ) and B(2, 3 y 2 ) AB = ((2) ( 4)) 2 + (( 3) (1)) 2 AB = 52 which your calculator will simplify to 2 13 (ii) Use the midpoint formula. x 1 y 1 x 2 A( 4,1 ) and B(2, 3 y 2 ) Midpoint of [AB] is ( ( 4)+(2) 2, (1)+( 3) ) = ( 2 2, ) = ( 1, 1) (iii) Use the slope of a line formula. x 1 y 1 x 2 A( 4,1 ) and B(2, 3 y 2 ) Slope of AB = ( 3) (1) (2) ( 4) = 2 3 (iv) 2x = 4 3y bring everything over onto the left hand side 2x + 3y 4 = 0 this is in the form ax + by + c = 0 a = the number in front of the x = 2. b = the number in front of the y = 3. So the slope is a b = 2 3. This is the same as the slope of AB from part (iii). As the two lines have the same slope, they are parallel. (v) Use the equation of a line formula. x 1 y 1 A( 4,1 ) and m = 2. 3 Equation of AB is y (1) = 2 (x ( 4)) multiply across by 3 3 3(y (1)) = 2(x ( 4)) 3(y 1) = 2(x + 4) now multiply out the brackets 3y 3 = 2x 8 now move everything onto the same side 3y 3 + 2x + 8 = 0 2x + 3y + 5 = 0 (vi) You need to find the slope of k. k is perpendicular to AB and AB has a slope of 2 3. Find the slope of k by turning the slope of AB upside-down and changing its sign. 2 becomes 3 and change the sign of the slope from negative to positive. 3 2 So the slope of k is 3. 2 Now use the equation of a line formula. x 1 y 1 C( 1, 1 ) from part (ii) and m = 3 2. The Dublin School of Grinds Page 136

139 Equation of AB is y ( 1) = 3 (x ( 1)) multiply across by 2 2 2(y ( 1)) = 3(x ( 1)) 2(y + 1) = 3(x + 1) now multiply out the brackets 2y + 2 = 3x + 3 now move everything onto the same side 2y + 2 3x 3 = 0 3x + 2y 1 = 0 or 3x 2y + 1 = 0 (vii) Put the x and y co-ordinates of (3,5) into the equation of k. 3(3) 2(5) + 1 = = 0 0 = 0 this is true! So the point D is on the line k. (viii) You are going to use the area of a triangle formula. But none of the corners (vertices) are at (0,0)! Move any one corner to (0,0) using a translation. Then remember to move the other two corners with the same translation. "add 4 to x, subtract 1 from y" A( 4,1) "add 4 to x, subtract 1 from y" (0,0) B(2, 3) (6, 4) "add 4 to x, subtract 1 from y" D(3,5) (7,4) Now put the two points that are not (0,0) into the formula. x 1 y 1 x 2 A(6, 4 ) and B(7,4 y 2 ) So the area of ABD = 1 (6)(4) (7)( 4) = = 1 52 = 26 units The Dublin School of Grinds Page 137

140 Question 3 (i) 3x + 2y + 18 = 0 this is in the form ax + by + c = 0 a = the number in front of the x = 3. b = the number in front of the y = 2. So the slope of l is a b = 3 2. (ii) You need to find the slope of k. k is perpendicular to l and l has a slope of 3 2. Find the slope of k by turning the slope of l upside-down and changing its sign. 3 becomes 2 and change the sign of the slope from negative to positive. 2 3 So the slope of k is 2. 3 Now use the equation of a line formula. x 1 y 1 (7,0 ) and m = 2. 3 Equation of k is y (0) = 2 (x (7)) multiply across by 3 3 3(y (0)) = 2(x (7)) 3(y) = 2(x 7) now multiply out the brackets 3y = 2x 14 now move everything onto the same side 3y 2x + 14 = 0 2x + 3y + 14 = 0 or 2x 3y 14 = 0 (iii) Solve the simultaneous equations. 3x + 2y + 18 = 0 2x 3y 14 = 0 Move the number across onto the right hand side in each equation. 3x + 2y = 18 2x 3y = 14 Multiply everything on both sides of the top equation by 3. Multiply everything on both sides of the bottom equation by 2. 3x + 2y = 18 becomes 9x + 6y = 54 2x 3y = 14 becomes 4x 6y = 28 add the equations together 13x + 6 = 26 x = 2 Now find the y co-ordinate by putting this x value into one of the equations. 3( 2) + 2y = y = 18 2y = 12 y = 6 So the point of intersection is ( 2, 6) The Dublin School of Grinds Page 138

141 Question 4 Slope Line p s q p and q both go downhill so the two negative slope values must belong to p and q. p is steeper than q so it must have the bigger number. Since 11 > 2 11, p must be and q must be r and s both go uphill so the two positive slope values must belong to r and s. s is steeper than r so it must have the bigger number. Since 5 2 > 1 4, s must be 5 2 and r must be r The Dublin School of Grinds Page 139

142 Question 5 (a) vertical change rise = 2 horizontal change run = 6 slope of k = rise run = 2 6 = 1 3. (ii) The line k passes through the point (0,1) which is marked on the diagram. Use the equation of a line formula. x 1 y 1 (0,1 ) and m = 1. 3 Equation of k is y (1) = 1 (x (0)) multiply across by 3 3 3(y (1)) = 1(x (0)) 3(y 1) = 1(x) now multiply out the brackets 3y 3 = x now move everything onto the same side 3y 3 x = 0 x + 3y 3 = 0 or x 3y + 3 = 0 (iii) You need to find two points on the line m, 2x + 3y 12 = 0. First find where the line cuts the x-axis by putting y = 0. 2x + 3(0) 12 = 0 2x 0 12 = 0 2x 12 = 0 2x = 12 x = 6 So the point is (6,0) Next find where the line cuts the y-axis by putting x = 0. 2(0) + 3y 12 = y 12 = 0 3y 12 = 0 3y = 12 y = 4 So the point is (0,4) Now plot these two points and join them together. The Dublin School of Grinds Page 140

143 (iv) From the graph, the two lines intersect at the point (3,2). (v) Solve the simultaneous equations. x 3y + 3 = 0 from part (ii) 2x + 3y 12 = 0 Move the number across onto the right hand side in each equation. x 3y = 3 2x + 3y = 12 add the equations together 3x 3y = 9 x 3y = 3 Now find the y co-ordinate by putting this x value into one of the equations. 2(3) + 3y = y = 12 3y = 6 y = 2 So the point of intersection is (3,2) same answer as part (iv) The Dublin School of Grinds Page 141

144 Question 6 (a) First you need to find the slope of AB. Use the slope of a line formula. x 1 y 1 x 2 A(6,1 ) and B(2, 1 y 2 ) Slope of AB = ( 1) (1) (2) (6) = 1 2. Now use the equation of a line formula. x 1 y 1 A(6,1 ) and m = 1. 2 Equation of AB is y (1) = 1 (x (6)) multiply across by 2 2 2(y (1)) = 1(x (6)) 2(y 1) = 1(x 6) now multiply out the brackets 2y 2 = x 6 now move everything onto the same side 2y 2 x + 6 = 0 x + 2y + 4 = 0 or x 2y 4 = 0 (b) Find where the line cuts the y-axis by putting x = 0. (0) 2y 4 = 0 2y 4 = 0 2y = 4 y = 2 So the point C is (0, 2) (c) First find AB. Use the distance formula. x 1 y 1 x 2 A(6,1 ) and B(2, 1 y 2 ) AB = ((2) (6)) 2 + (( 1) (1)) 2 AB = 20 which your calculator will simplify to 2 5 Now find AC x 1 y 1 x 2 y 2 A(6,1 ) and C(0, 2 ) from part (b) AC = ((0) (6)) 2 + (( 2) (1)) 2 AC = 45 which your calculator will simplify to 3 5 So AB = 2 5 = 2 when you cancel 5 above and below AC The Dublin School of Grinds Page 142

145 Question 7 a) (i) x 1 y 1 x 2 y 2 A( 9, 3) B( 4, 3) AB = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 = (( 4) ( 9)) 2 + ((3) (3)) 2 = ( 4 + 9) 2 = (5) 2 = 5 (ii) Remember we must move one vertex to (0, 0) and but the pther two vertices must also undergo the same translation. Let s move B to (0, 0) since it s the closest Next, A: Finally,C: x: +4 ( 4, 3) (0,0) y: 3 x: +4 ( 9, 3) (5,0) y: 3 x: +4 ( 4, 10) (0,7) y: 3 x 1 y 1 x 2 y 2 (5, 0) (0, 7) Area of a triangle = 1 2 x 1y 2 x 2 y 1 = 1 (5)(7) (0)(0) 2 b) (i) = = 35 2 units2 The Dublin School of Grinds Page 143

146 (ii) Z = ( 5, 1) Reason: AB = XY = 5units BC = YZ = 7 units ABC = XYZ both 90 By the Rules for congruence (SAS) the triangles ABC and XYZ are congruent. Question 8 (a) (i) Slope of p = 1 (b) (c) (ii) Point = (1, 5) m = 1 Point = (0, 0) m = -1 y y 1 = m(x x 1 ) y 5 = 1(x 1) y 5 = x 1 x y + 4 = 0 q p => slope of q = 1 y y 1 = m(x x 1 ) y 0 = 1(x 0) y = x x + y = 0 ACO = BCO both 90 as q p AO = BO given CO = CO common => RHS => OCA and OBC are congruent. The Dublin School of Grinds Page 144

147 Co-ordinate geometry of the circle Coordinate geometry of the circle is worth 4% to 6% of the Leaving Cert. It appears on Paper Equation of a circle with centre (0, 0) The Examiner can ask you two things about a circle with centre (0, 0): 1. To find the equation of a circle. 2. To find the radius of a circle. The equation of a circle with centre (0,0) and radius r is NOTE: This is not in the log tables so you need to learn it. x 2 + y 2 = r 2 Example 1 Find the equation of the circle with centre (0, 0), which has a radius of 5. Solution Substitute r = 5 into this equation: x 2 + y 2 = 5 2 x 2 + y 2 = 25 In the Leaving Cert exam you can be asked to find the radius. Example 2 Find the equation of the circle c whose centre is (0, 0) and which contains the point (-2, 5). Solution The radius is the distance from the centre (0, 0) to the point (-2, 5). r = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 = (( 2) 0) 2 + (5 0) 2 = ( 2) 2 + (5) 2 = = 29 Now we substitute r into the equation of a circle: x 2 + y 2 = r 2 x 2 + y 2 = ( 29) 2 x 2 + y 2 = 29 (length formula from page 18 of log tables) Question 1.1 Find the equation of the circle with centre (0, 0) and which contains the point (3, 1). The Dublin School of Grinds Page 145

148 Example 3 Find the radius of the circle x 2 + y 2 = 25. Solution Compare x 2 + y 2 = r 2 with x 2 + y 2 = 25 => r 2 = 25 r = 5 Question 1.2 Find the radius of the following circles: i) x 2 + y 2 = 100 ii) x 2 + y 2 = 12 Example 4 The points (6, 1) and (-6, -1) are the end points of the diameter of a circle. Find i) the coordinates of the centre of the circle. ii) the length of the radius iii) the equation of the circle. Solution i) The centre is the midpoint of the diameter: midpoint = ( x 1 + x 2 2 = ( 6 6, y 1 + y 2 ) 2 2, ) = (0, 0) (midpoint formula from page 18 of log tables) ii) The length of the radius is the distance from the centre (0, 0) to one of the end points on the diameter (6, 1): r = (6 0) 2 + (1 0) 2 = = 37 iii) Now we substitute r into the equation of a circle: x 2 + y 2 = r 2 x 2 + y 2 = ( 37) 2 x 2 + y 2 = 37 The Dublin School of Grinds Page 146

149 Question 1.3 The points (4, 3) and (-4, -3) are the end points of the diameter of a circle. Find i) the coordinates of the centre of the circle. ii) the length of the radius iii) the equation of the circle. The Dublin School of Grinds Page 147

150 2. Point inside, on or outside a circle with centre (0, 0) The Examiner can ask you to verify if a point is inside, on or outside a circle. Example 1 Are the points i) (1, 3) ii) (3, 2) iii) (5, 4) inside, outside or on the circle x 2 + y 2 = 13? Solution i) Substitute 1 in for x and 3 for y into the equation x 2 + y 2 = 13: (1) 2 + ( 3) 2 = < < 13 => the point is inside the circle ii) Substitute 3 in for x and 2 for y into the equation x 2 + y 2 = 13: (3) 2 + ( 2) 2 = = = 13 => the point is on the circle iii) Substitute 5 in for x and 4 for y into the equation x 2 + y 2 = 13: (5) 2 + ( 4) 2 = > > 13 => the point is outside the circle Question 2.1 Check if the point (-5, 1) is on the circle x 2 + y 2 = 25. The Dublin School of Grinds Page 148

151 Question 2.2 Show that the point (3, 2) is on the circle x 2 + y 2 = 13 and hence draw the graph of the circle. The Dublin School of Grinds Page 149

152 3. The Equation of a circle with the centre not at (0, 0) The Leaving Cert syllabus requires you to know how to find the equation of the circle when the centre is not (0, 0). The equation of a circle with centre (h, k) and radius length r is (x h) 2 + (y k) 2 = r 2 Note: This formula is on page 19 of the log tables. Example 1 Find the equation of the circle centre ( 3, 5) and radius 4. Solution h = 3 k = 5 r = 4 (x h) 2 + (y k) 2 = r 2 (x ( 3)) 2 + (y (5)) 2 = (4) 2 (x + 3) 2 + (y 5) 2 = 16 Question 3.1 Find the equation of the circle centre ( 4, 3) and radius 17. The Examiner may not give you the radius but he will give you the information to find it. Example 2 The centre of the circle (2, 4) and it contains the point ( 1, 3). Find the equation of the circle. r = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 r = ( 1 2) 2 + (3 4) 2 = ( 3) 2 + ( 1) 2 = 10 h = 2 k = 4 r = 10 (x h) 2 + (y k) 2 = r 2 (x (2)) 2 + (y (4)) 2 = ( 10 ) 2 (x 2) 2 + (y 4) 2 = 10 Question 3.2 Find the equation of a circle with centre (1, 2) and containing the point (2, 5). The Dublin School of Grinds Page 150

153 Question 3.3 A line segment joining (3, 5) and ( 1, 1), is the diameter of a circle. Find the equation of the circle. You may be asked in the Leaving Cert Exam to find the centre and the radius of a circle when you are given the equation. Example 3 Find the centre of the circle and the length of the radius (x 3) 2 + (y 5) 2 = 4 Solution Comparing equations: (x h) 2 + (y k) 2 = r 2 and (x 3) 2 + (y 5) 2 = 4 => h = 3 k = 5 and r 2 = 4 => r = 2 => the centre (3, 5) and radius = 2 Be careful if the sign in the brackets is positive: Example 4 Find the centre of the circle and the length of the radius (x + 2) 2 + (y + 4) 2 = 16 Solution Comparing equations: (x h) 2 + (y k) 2 = r 2 and (x + 2) 2 + (y + 4) 2 = 16 => h = 2, k = 4 and r 2 = 16 => r = 4 => the centre ( 2, 4) and radius = 4 Question 3.4 Find the centre and radius of the following circles: i) (x 1) 2 + (y + 5) 2 = 36 ii) (x + 2) 2 + (y 5) 2 = 25 The Dublin School of Grinds Page 151

154 4. Point in, on or outside a circle with centre not (0, 0) We follow the same procedure as in Section 2. Example 1 Determine if the point (3, 4) is inside, on or outside the circle (x + 1) 2 + (y + 4) 2 = 1. Solution (x + 1) 2 + (y + 4) 2 = (3 + 1) 2 + ( 4 + 4) 2 = (4) 2 + (0) 2 = > 1 => (3, 4)is outside the circle Question 4.1 Determine if the point ( 2, 1) is inside, on or outside the circle (x + 3) 2 + (y 1) 2 = 16. Question 4.2 Determine if the point (4, 3) is inside, on or outside the circle (x 2) 2 + (y + 1) 2 = 20. The Dublin School of Grinds Page 152

155 5. Point of intersection between a circle and a line The Examiner can ask you to find the point of intersection between a line and a circle and there are three possible outcomes. Firstly we will look at a problem with two points of intersection. To solve these problems we use simultaneous equations. Example 1 Find the points of intersection between the line x y = 3 and the circle x 2 + y 2 = 17. Solution First, number the equations (1) x y = 3 (2) x 2 + y 2 = 17 Rearrange (1) to find y on its own. Next, substitute x 3 into equation (2) for y. x y = 3 y = x + 3 y = x 3 x 2 + y 2 = 17 x 2 + (x 3) 2 = 17 x 2 + (x 3)(x 3) = 17 x 2 + x 2 3x 3x + 9 = 17 2x 2 6x + 9 = 17 Rearrange the equation to have everything on the left. 2x 2 6x = 0 2x 2 6x 8 = 0 a = 2, b = 6, c = 8 x = b ± b2 4ac 2a x = ( 6) ± ( 6)2 4(2)( 8) 2(2) 6 ± x = 6 ± x = 6 ± 10 4 x = x = x = 16 or x = or x = x = 4 or x = 1 The Dublin School of Grinds Page 153

156 Finally we substitute the x values into equation (1): x = 4 1) x y = 3 (4) y = 3 y = 3 4 y = 1 y = 1 => x = 4 & y = 1 x = 1 1) x y = 3 ( 1) y = 3 y = y = 4 y = 4 => x = 1 & y = 4 The points of intersection are (4, 1) and ( 1, 4). Question 5.1 Find the points of intersection between the line x + 5y + 13 = 0 and the circle x 2 + y 2 = 13. The Dublin School of Grinds Page 154

157 Example 2 Find the point of intersection between the line 3x + y 10 = 0 and the circle x 2 + y 2 = 10 and investigate if this line is a tangent. Solution First, number the equations 1) 3x + y 10 = 0 2) x 2 + y 2 = 10 Rearrange 1) to find y on its own. Next, substitute x + 1 into equation 2) for y. Rearrange the equation to have everything on the left. Now solve the quadratic equation Now substitute x = 3 back into equation 1) 3x + y 10 = 0 y = 10 3x x 2 + y 2 = 10 x 2 + (10 3x) 2 = 10 x 2 + (10 3x)(10 3x) = 10 x x 30x + 9x 2 = 10 10x 2 60x = 10 10x 2 60x = 0 10x 2 60x + 90 = 0 a = 10, b = 60, c = 90 x = b ± b2 4ac 2a x = ( 60) ± ( 60)2 4(10)(90) 2(10) x = 60 ± 0 20 x = x = 3 3x + y 10 = 0 3(3) + y 10 = y 10 = 0 y 1 = 0 y = 1 The point of intersection is (3, 1). Since there is only one point of intersection the line must be a tangent. The Dublin School of Grinds Page 155

158 Question 5.2 Find the point of intersection between the line 2x y = 5 and the circle x 2 + y 2 = 5 and investigate if this line is a tangent. The Dublin School of Grinds Page 156

159 6. Finding the equation of a tangent The Examiner can ask you to find the equation of a tangent to a circle. To find the tangent follow these 4 steps: Step 1: Find the centre of the circle. Step 2: Find the slope of the radius to the point on the circle. Step 3 Find the slope of the tangent, (it will be perpendicular to the slope in step 2.) Step 4: Use the formula y y 1 = m(x x 1 ) from page 18 of the log tables. Example 1 Find the equation of a tangent to the circle (x 4) 2 + (y + 2) 2 = 13 at the point (6, 5) on the circle. Solution Step 1: (x 4) 2 + (y + 2) 2 = 13 centre = (4, 2) Step 2: Slope of the radius using the centre (6, 5) and the point (4, 2): slope of radius = y 2 y 1 x 2 x 1 = = 3 2 Step 3: The slope of the tangent is perpendicular to the slope in Step 2: Slope of tangent = 2 3 Step 4: x1 = 6 y1 = 5 m = 2 3 (y y 1 ) = m(x x 1 ) (y ( 5)) = 2 (x 6) 3 (y + 5) = 2 (x 6) 3 3(y + 5) = 2(x 6) 3y + 15 = 2x 12 2x + 3y = 0 2x + 3y + 27 = 0 2x 3y 27 = 0 The Dublin School of Grinds Page 157

160 Question 6.1 Find the equation of a tangent to the circle (x + 2) 2 + (y 3) 2 = 29 at the point (3, 5) on the circle. The Dublin School of Grinds Page 158

161 7. Finding the points of intersection with the axes There are two different ways the Examiner can ask where a circle crosses the axes: 1. When the circle crosses the x-axis: y = 0 2. When the circle crosses the y-axis: x = 0 Example 1 Find the coordinates of the points where the circle (x 2) 2 + (y 3) 2 = 25 intersects with the x-axis. Solution When the circle intersects with the x-axis: y = 0. We substitute y = 0 into the equation of the circle: (x 2) 2 + (0 3) 2 = 25 (x 2)(x 2) + ( 3) 2 = 25 x 2 2x 2x = 25 x 2 4x + 13 = 25 x 2 4x = 0 x 2 4x 12 = 0 Now solve the quadratic equation using the formula from the cover of the log tables: a = 1, b = 4, c = 12 x = b ± b2 4ac 2a x = ( 4) ± ( 4)2 4(1)( 12) 2(1) x = x = x = 12 2 The circle intersects the x axis at (6, 0) and ( 2, 0) 4 ± x = 4 ± 64 2 x = 4 ± 8 2 or x = or x = 4 2 x = 6 or x = 2 The Dublin School of Grinds Page 159

162 Question 7.1 Find the coordinates of the points where the circle (x 2) 2 + (y + 3) 2 = 20 intersects with the y-axis. The Dublin School of Grinds Page 160

163 Question 7.2 Verify that the x-axis is a tangent to the circle (x + 2) 2 + (y 4) 2 = 16. Hint: If there is one point of intersection it is a tangent. The Dublin School of Grinds Page 161

164 8. Past and probable exam questions Question 1 The Dublin School of Grinds Page 162

165 Question 2 The Dublin School of Grinds Page 163

166 Question 3 The Dublin School of Grinds Page 164

167 The Dublin School of Grinds Page 165

168 Question 4 The Dublin School of Grinds Page 166

169 The Dublin School of Grinds Page 167

170 Question 5 The Dublin School of Grinds Page 168

171 Question 6 The Dublin School of Grinds Page 169

172 The Dublin School of Grinds Page 170

173 Question 7 The Dublin School of Grinds Page 171

174 Question 8 The Dublin School of Grinds Page 172

175 The Dublin School of Grinds Page 173

176 Question 9 The Dublin School of Grinds Page 174

177 Question 10 The Dublin School of Grinds Page 175

178 9. Solutions to Co-ordinate geometry of the circle Question 1.1 The radius is the distance from the centre (0, 0) to the point (3, 1). r = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 = ((3) 0) 2 + (1 0) 2 = (3) 2 + (1) 2 = = 10 Now we substitute r into the equation of a circle: x 2 + y 2 = r 2 x 2 + y 2 = ( 10) 2 x 2 + y 2 = 10 Question 1.2 i) ii) Compare x 2 + y 2 = r 2 with x 2 + y 2 = 100 => r 2 = 100 r = 10 Compare x 2 + y 2 = r 2 with x 2 + y 2 = 12 => r 2 = 12 r = 12 Question 1.3 i) The centre is the midpoint of the diameter: midpoint = ( x 1 + x 2 2 = ( 4 4, y 1 + y 2 ) 2 2, ) = (0, 0) ii) The length of the radius is the distance from the centre (0, 0) to one of the end points on the diameter (4, 3): r = (4 0) 2 + (3 0) 2 = = 25 = 5 iii) Now we substitute r into the equation of a circle: x 2 + y 2 = r 2 x 2 + y 2 = (5) 2 x 2 + y 2 = 25 The Dublin School of Grinds Page 176

179 Question 2.1 Substitute 5 in for x and 1 for y into the equation x 2 + y 2 = 25 ( 5) 2 + (1) 2 = > > 25 => the point is outside the circle Question 2.2 Substitute 3 in for x and 2 for y into the equation x 2 + y 2 = 13 (3) 2 + (2) 2 = = = 13 => the point is on the circle Question 3.1 h = 4 k = 3 r = 17 (x h) 2 + (y k) 2 = r 2 (x ( 4)) 2 + (y ( 3)) 2 = ( 17) 2 (x + 4) 2 + (y + 3) 2 = 17 Question 3.2 The centre of the circle (1, 2) and it contains the point (2, 5). Find the equation of the circle. r = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 h = 1 k = 2 r = 10 r = (2 1) 2 + (5 2) 2 = (1) 2 + (3) 2 = 10 (x h) 2 + (y k) 2 = r 2 (x (1)) 2 + (y (2)) 2 = ( 10 ) 2 (x 1) 2 + (y 2) 2 = 10 Question 3.3 The centre is the midpoint of the diameter. Using points (3, 5) and ( 1, 1): midpoint = ( x 1 + x 2 2 = ( 3 1, y 1 + y 2 ) 2 2, ) = (1, 3) The centre of the circle (1, 3) and it contains the point (3, 5). Find the equation of the circle. r = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 h = 1 k = 3 r = 8 r = (3 1) 2 + (5 3) 2 = (2) 2 + (2) 2 = 8 (x h) 2 + (y k) 2 = r 2 (x (1)) 2 + (y (3)) 2 = ( 8 ) 2 (x 1) 2 + (y 3) 2 = 8 The Dublin School of Grinds Page 177

180 Question 3.4 i) Comparing equations: ii) (x h) 2 + (y k) 2 = r 2 and (x 1) 2 + (y + 5) 2 = 36 => h = 1 k = 5 and r 2 = 36 => r = 6 => the centre (1, 5) and radius = 6 Comparing equations: (x h) 2 + (y k) 2 = r 2 and (x + 2) 2 + (y 5) 2 => h = 2 k = 5 and r 2 = 25 => r = 5 => the centre ( 2, 5) and radius = 5 Question 4.1 (x + 3) 2 + (y 1) 2 = ( 2 + 3) 2 + (1 1) 2 = (1) 2 + (0) 2 = 1 1 < 16 => ( 2, 1)is inside the circle Question 4.2 (x 2) 2 + (y + 1) 2 = (4 2) 2 + (3 + 1) 2 = (2) 2 + (4) 2 = = 20 => (4, 3)is on the circle Question 5.1 First, number the equations (1) x + 5y + 13 = 0 (2) x 2 + y 2 = 13 Rearrange (1) to find y on its own. Next, substitute 5y 13 into equation (2) for y. Now solve the quadratic equation y = x + 5y + 13 = 0 x = 5y 13 x 2 + y 2 = 13 ( 5y 13) 2 + y 2 = 13 ( 5y 13)( 5y 13) + y 2 = 13 25y y y 2 = 13 26y y = 0 a = 26, b = 130, c = 156 x = b ± b2 4ac 2a y = (130) ± (130)2 4(26)(156) 2(26) 130 ± y = ± 676 y = y = y = ± or y = or y = y = 2 or y = 3 The Dublin School of Grinds Page 178

181 Finally we substitute the y values into the equation 1 y = 3 2) x + 5y + 13 = 0 x + 5( 3) + 13 = 0 x = 2 => x = 2 & y = 3 y = 2 2) x + 5y + 13 = 0 x + 5( 2) + 13 = 0 x = 3 => x = 3 & y = 2 The points of intersection are (2, 3) and ( 3, 2). Question 5.2 First, number the equations 1) 2x y = 5 2) x 2 + y 2 = 5 Rearrange 1) to find y on its own. Next, substitute 2x 5 into equation 2) for y. Rearrange the equation to have everything on the left. Now solve the quadratic equation Now substitute x = 2 back into equation 1 2x y = 5 y = 2x 5 x 2 + y 2 = 5 x 2 + (2x 5) 2 = 5 x 2 + (2x 5)(2x 5) = 5 x 2 + 4x 2 10x 10x + 25 = 5 5x 2 20x + 25 = 5 5x 2 20x = 0 5x 2 20x + 20 = 0 a = 5, b = 20, c = 20 x = b ± b2 4ac 2a x = ( 20) ± ( 20)2 4(50)(200) 2(5) x = 20 ± 0 10 x = x = 2 2x y = 5 2(2) y = 5 4 y = 5 y = 1 y = 1 The point of intersection is (2, 1). Since there is only one point of intersection the line must be a tangent. The Dublin School of Grinds Page 179

182 Question 6.1 Step 1: (x + 2) 2 + (y 3) 2 = 29 centre = ( 2, 3) Step 2: Slope of the radius using the centre ( 2, 3) and the point (3, 5): slope of radius = y 2 y 1 x 2 x 1 = = 2 5 Step 3: The slope of the tangent is perpendicular to the slope in Step 2: Slope of tangent = 5 2 Step 4: x1 = 3 y1 = 5 m = 5 2 (y y 1 ) = m(x x 1 ) (y 5) = 5 (x 3) 2 2(y 5) = 5(x 3) 2y 10 = 5x x + 2y = 0 5x + 2y 25 = 0 Question 7.1 When the circle intersects with the y-axis: x = 0. We substitute x = 0 into the equation of the circle: (0 2) 2 + (y + 3) 2 = 20 ( 2) 2 + (y + 3)(y + 3) = y 2 + 3y + 3y + 9 = 20 y 2 + 6y + 13 = 20 y 2 + 6y = 0 y 2 + 6y 7 = 0 Now solve the quadratic equation using the formula from the cover of the log tables: a = 1, b = 6, c = 7 y = b ± b2 4ac 2a y = (6) ± (6)2 4(1)( 7) 2(1) y = y = y = ± ± 64 y = 2 y = 6 ± 8 2 or y = or y = 14 2 y = 1 or y = 7 The circle intersects the x axis at (0, 1) and (0, 7) The Dublin School of Grinds Page 180

183 Question 7.2 When the circle intersects with the x-axis: y = 0. We substitute y = 0 into the equation of the circle: (x + 2) 2 + (0 4) 2 = 16 (x + 2)(x + 2) + ( 4) 2 = 16 x 2 + 2x + 2x = 16 x 2 + 4x + 20 = 16 x 2 + 4x = 0 x 2 + 4x + 4 = 0 Now solve the quadratic equation using the formula from the cover of the log tables: a = 1, b = 4, c = 4 x = b ± b2 4ac 2a x = (4) ± (4)2 4(1)(4) 2(1) x = 4 ± 0 2 x = 4 2 x = 2 The circle intersects the x axis at ( 2, 0) only so therefore it is a tangent. The Dublin School of Grinds Page 181

184 Past and probable exam question solutions Question 1 a) b) The centre of the circle (-2, -1) and it contains the point (3, 1). Find the equation of the circle. r = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 r = (3 + 2) 2 + (1 + 1) 2 = (5) 2 + (4) 2 = 29 h = -2 k = -1 r = 29 (x h) 2 + (y k) 2 = r 2 (x ( 2)) 2 + (y ( 1)) 2 = ( 29 ) 2 (x + 2) 2 + (y + 1) 2 = 29 c) Find the slope of PR using P(-2, -1) and Q(3, 1) m 1 = y 2 y 1 x 2 x 1 = Find the slope of QR using Q(3, 1) and R(1, 6) = 2 5 m 2 = y 2 y 1 x 2 x 1 = = 5 2 m 1 m 2 = 1 => PR and QR are perpendicular. Since PR is perpendicular to QR and QR touches the circle at Q it is a tangent. The Dublin School of Grinds Page 182

185 Question 2 a) The radius is the distance from the centre (0, 0) to the point (8, 6). r = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 = (8 0) 2 + (6 0) 2 = (8) 2 + (6) 2 = 100 = 10 Now we substitute r into the equation of a circle: x 2 + y 2 = r 2 x 2 + y 2 = (10) 2 x 2 + y 2 = 100 b) The centre is the midpoint of the diameter OA: midpoint = ( x 1 + x 2 2, y 1 + y 2 ) 2 = ( , ) 2 = (4, 3) The length of the radius is the distance from the centre (4, 3) to one of the end points on the diameter (0, 0): r = (4 0) 2 + (3 0) 2 = = 25 = 5 Now we substitute r into the equation of a circle: h = 4 k = 3 r = 5 (x h) 2 + (y k) 2 = r 2 (x 4) 2 + (y 3) 2 = (5) 2 (x 4) 2 + (y 3) 2 = 25 Solution continued on the next page. The Dublin School of Grinds Page 183

186 c) When the circle intersects with the x-axis: y = 0. We substitute y = 0 into the equation of the circle: (x 4) 2 + (0 3) 2 = 25 (x 4)(x 4) + ( 3) 2 = 25 x 2 4x 4x = 25 x 2 8x + 25 = 25 x 2 8x = 0 x 2 8x = 0 Now solve the quadratic equation using the formula from the cover of the log tables: a = 1, b = 8, c = 0 x = b ± b2 4ac 2a x = ( 8) ± ( 8)2 4(1)(0) 2(1) x = x = 16 2 x = 8 ± 64 2 x = 8 ± 8 2 or x = or x = 0 2 x = 8 or x = 0 The circle intersects the x axis at (0, 0) and P (8, 0) The Dublin School of Grinds Page 184

187 Question 3 a) The radius is the distance from the centre (0, 0) to the point (5, 0). r = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 = (5 0) 2 + (0 0) 2 = (5) 2 = 5 Now we substitute r into the equation of a circle: x 2 + y 2 = r 2 x 2 + y 2 = (5) 2 x 2 + y 2 = 25 b) Substitute -3 in for x and 4 for y into the equation x 2 + y 2 = 25: ( 3) 2 + ( 4) 2 = = = 25 => the point is on the circle c) We can find the centre of c1 using a translation: The centre of c2is ( 6,8) and the radius. h = -6 k = 8 r = 5 x: 3 (0,0) ( 3,4) y: +4 x: 3 ( 3,4) ( 6,8) y: +4 (x h) 2 + (y k) 2 = r 2 (x + 6) 2 + (y 8) 2 = (5) 2 (x + 6) 2 + (y 8) 2 = 25 The Dublin School of Grinds Page 185

188 d) Slope of the radius using the centre (0, 0) and the point P(-3, 4): slope of radius = y 2 y 1 x 2 x 1 = = 4 3 The slope of the tangent is perpendicular to the slope of the radius. Slope of tangent = 3 4 x1 = -3 y1 = 4 m = 3 4 (y y 1 ) = m(x x 1 ) (y 4) = 3 (x ( 3)) 4 (y 4) = 3 (x + 3) 4 4(y 4) = 3(x + 3) 4y 16 = 3x + 9 3x + 4y 16 9 = 0 3x + 4y 25 = 0 3x 4y + 25 = 0 The Dublin School of Grinds Page 186

189 Question 4 a) The circle has centre (0, 0). b) i) To find the radius: ii) Compare x 2 + y 2 = r 2 with x 2 + y 2 = 81 => r 2 = 81 r = 9 (x 3) 2 + (y + 1) 2 = (7 3) 2 + ( 2 + 1) 2 = (4) 2 + ( 1) 2 = = 17 => (7, 2)is on the circle iii) When the circle intersects with the x-axis: y = 0. We substitute y = 0 into the equation of the circle: (x 3) 2 + (0 + 1) 2 = 17 (x 3)(x 3) + (1) 2 = 17 x 2 3x 3x = 17 x 2 6x + 10 = 17 x 2 6x = 0 x 2 6x 7 = 0 Continued on the next page. The Dublin School of Grinds Page 187

190 Now solve the quadratic equation using the formula from the cover of the log tables: a = 1, b = 6, c = 7 x = b ± b2 4ac 2a x = ( 6) ± ( 6)2 4(1)( 7) 2(1) x = x = x = 14 6 ± x = 6 ± 64 2 x = 6 ± 8 2 or x = or x = x = 7 or x = 1 The circle intersects the x axis at (7, 0) and ( 1, 0) c) i) The centre of the circle is the midpoint of AC Using points (-1, 2) and (3, 6): midpoint = ( x 1 + x 2 2 = ( 1 + 3, y 1 + y 2 ) 2, ) 2 = (1, 2) => the centre of s is (1, 2) ii) To find the equation of s we need the radius. To find the radius we need to find a point on the circle. We can use the midpoint of AD. Using the points (-1, 2) and (5, 0): midpoint = ( x 1 + x 2 2 = ( 1 + 5, y 1 + y 2 ) 2, ) = (2, 1) => (2,1)is a point on the circle The centre of the circle (1, -2) and it contains the point (2, 1). Find the equation of the circle. h = 1 k = -2 r = 10 r = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 r = (2 1) 2 + (1 + 2) 2 = (1) 2 + (3) 2 = 10 (x h) 2 + (y k) 2 = r 2 (x (1)) 2 + (y ( 2)) 2 = ( 10 ) 2 (x 1) 2 + (y + 2) 2 = 10 iii) Compare the circle given to you to the formula given to you in the Formulae and log booklet: (x h) 2 + (y k) 2 = r 2 and (x + 4) 2 + y 2 = 4 => h = 4 k = 0 => the centre ( 4, 0) To find the value of a p and q find the translations of the centre image of s to the centre s: x: 5 (1, 2) ( 4,0) y: +2 x: 5 (p, q) (6,5) y: +2 p 5 = 6 and q + 2 = 5 => p = 11 and q = 3 The Dublin School of Grinds Page 188

191 Question 5 a) the diameter is 8 units so the radius is 4 units. b) Substitute r = 4 into this equation: x 2 + y 2 = 4 2 x 2 + y 2 = 16 c) Substitute 3 in for x and 2 for y into the equation x 2 + y 2 = 16: (3) 2 + (2) 2 = < < 16 => the point is inside the circle Substitute 3 in for x and 3 for y into the equation x 2 + y 2 = 16: (3) 2 + (3) 2 = > > 16 => the point is outside the circle d) The new circle c2 only touches the circle c1 at one point. The radius of c2 is 3 which can be seen in the diagram. h = 0 k = 1 r = 3 (x h) 2 + (y k) 2 = r 2 (x (0)) 2 + (y (1)) 2 = (3) 2 x 2 + (y 1) 2 = 9 The Dublin School of Grinds Page 189

192 Question 6 a) i) The centre of the circle (0, 0) and it contains the point (5, -12). Find the equation of the circle. r = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 r = (5 0) 2 + ( 12 0) 2 = (5) 2 + ( 12) 2 = 169 = 13 ii) Now we substitute r into the equation of a circle: x 2 + y 2 = r 2 x 2 + y 2 = (13) 2 x 2 + y 2 = 169 b) i) First, number the equations 1) x 4y 17 = 0 2) x 2 + y 2 = 17 Rearrange 1) to find y on its own. x 4y 17 = 0 x = 4y + 17 Next, substitute (4y + 17) into equation 2) for x. x 2 + y 2 = 17 (4y + 17) 2 + y 2 = 17 (4y + 17)(4y + 17) + y 2 = 17 16y y + 68y y 2 = 17 17y y = 17 Rearrange the equation to have everything on the left. 17y y = 0 17y y = 0 Now solve the quadratic equation a = 17, b = 136, c = 272 y = b ± b2 4ac 2a y = (136) ± (136)2 4(17)(272) 2(17) y = 136 ± 0 34 y = x = 4 Now substitute y = -4 back into equation 1 x 4y 17 = 0 x 4( 4) 17 = 0 x = 0 x 1 = 0 => x = 1 The co-ordinates of T are (1, 4). ii) To find to other end of the diameter we can use the translation from T to the centre; (1, 4) (0,0) x: 1 (0,0) ( 1,4) y: +4 x: 1 y: +4 => The other end point is ( 1,4) The Dublin School of Grinds Page 190

193 c) i) ii) Substitute 6 in for x and h in for y: (x h) 2 + (y k) 2 = r 2 and x 2 + (y 7) 2 = 100 => h = 0 k = 7 and r 2 = 100 => r = 10 => the centre (0, 7) and radius = (h 7) 2 = h 2 7h 7h + 49 = 100 h 2 14h = 0 h 2 14h 15 = 0 Now solve the quadratic equation using the formula from the cover of the log tables: a = 1, b = 14, c = 15 h = b ± b2 4ac 2a h = ( 14) ± ( 14)2 4(1)( 15) 2(1) 14 ± 256 h = 2 14 ± 16 h = h = 2 or h = h = 30 2 or h = 2 2 h = 15 or h = The Dublin School of Grinds Page 191

194 Question 7 a) i) The centre of the circle (0, 0) and it contains the point (3, 4). Find the equation of the circle. r = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 r = (3 0) 2 + (4 0) 2 = (3) 2 + (4) 2 = 25 = 5 Now we substitute r into the equation of a circle: x 2 + y 2 = r 2 x 2 + y 2 = (5) 2 x 2 + y 2 = 25 ii) When the circle crosses the y-axis x = 0 x 2 + y 2 = 25 (0) 2 + y 2 = 25 y 2 = 25 y = 25 y = ±5 => the circle crosses the y axis at (0, 5)and (0, 5) b) The radius of the circle in the diagram is 2. h = 2 k = 4 r = 2 (x h) 2 + (y k) 2 = r 2 (x (2)) 2 + (y (4)) 2 = (2) 2 (x 2) 2 + (y 4) 2 = 4 The Dublin School of Grinds Page 192

195 Question 8 a) i) Compare x 2 + y 2 = r 2 with x 2 + y 2 = 25 => r 2 = 25 r = 5 ii) Substitute 4 in for x and 3 for y into the equation x 2 + y 2 = 25: (4) 2 + ( 3) 2 = = = 25 => the point is on the circle iii) We know the centre so we will start by finding the slope using the centre (0, 0) and the point (4,-3) slope of radius = y 2 y 1 x 2 x 1 = = 3 4 The slope of the tangent is perpendicular to the slope of the radius. Slope of tangent = 4 3 x1 = 4 y1 = 3 m = 3 4 (y y 1 ) = m(x x 1 ) (y ( 3)) = 4 (x 4) 3 (y + 3) = 4 (x 4) 3 3(y + 3) = 4(x 4) 3y + 9 = 4x 16 4x + 3y = 0 4x + 3y + 25 = 0 4x 3y 25 = 0 iv) v) The two possible tangents parallel to the x-axis are where C crosses the y-axis are: y = 5 and y = 5 The Dublin School of Grinds Page 193

196 b) i) The centre of the circle (1,-6) and it contains the point (9, 0). Find the equation of the circle. r = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 r = (9 1) 2 + (0 + 6) 2 = (8) 2 + (6) 2 = 100 = 10 ii) h = 1 k = -6 r = 10 (x h) 2 + (y k) 2 = r 2 (x (1)) 2 + (y ( 6)) 2 = (10) 2 (x 1) 2 + (y + 6) 2 = 100 iii )you must use translations to find t. (9,0) (1, 6) (1, 6) ( 7, 12) => the coordinates of t are ( 7, 12) s will have the same x value as t because it is directly above it. s will have the same y value as r because it is horizontally across from it. => the coordinates of s are ( 7,0) u will have the same x value as r because it is directly below it. u will have the same y value as t because it is horizontally across from it. => the coordinates of u are (9, 12) x: 8 y: 6 x: 8 y: 6 The Dublin School of Grinds Page 194

197 Question 9 a) (i) Comparing equations: (x h) 2 + (y k) 2 = r 2 and (x + 2) 2 + (y 3) 2 = 100 => h = 2, k = 3 and r 2 = 100 => r = 10 => the centre is ( 2 3) and radius = 10 (ii) (x + 2) 2 + (y 3) 2 = (( 8) + 2) 2 + ((11) 3) 2 = ( 6) 2 + (8) 2 = = 100 => 100 = 100 => ( 8, 11) is on the circle b) (i) x 1 y 1 x 2 y 2 A( 2, 3) P( 8, 11) m = y 2 y 1 x 2 x m = ( 8) ( 2) = 8 6 = 4 3 (ii) The slope of the tangent is 3 (it is perpendicular to the slope of the radius) 4 Point ( 8, 11) y y 1 = m(x x 1 ) y 11 = 3 (x ( 8)) 4 Equations of t: y 11 = 3 (x + 8) 4 4(y 11) = 3(x + 8) 4y 44 = 3x = 3x 4y = 3x 4y + 68 => 3x 4y + 68 = 0 c) We can find the coordinates of Q using translations: A P Q x: + 6 x: +6 The slope is the same as t: slope = 3 4 point = (4, 5) ( 8, 11) ( 2, 3) (4, 5) y: 8 y: 8 y y 1 = m(x x 1 ) y ( 5) = 3 (x 4) 4 Equations of k: y + 5 = 3 (x 4) 4 4(y + 5) = 3(x 4) 4y + 20 = 3x 12 0 = 3x 4y = 3x 4y 32 => 3x 4y 32 = 0 The Dublin School of Grinds Page 195

198 Question 10 (a) Centre: (0, 0) radius = 5 (b) x 2 + y 2 = 25 ( 4) 2 + (3) 2 = = 25 (c) Find the distance from (-9, 3) to (3, 4) to find the radius: d = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 d = (3 ( 4)) 2 + (4 3) 2 Centre (-4, 3) => h = 4 => k = 3 = (7) 2 + (1) 2 = 50 => radius = 50 (x h) 2 + (y k) 2 = r 2 (x + 4) 2 + (y 3) 2 = 50 The Dublin School of Grinds Page 196

199 Geometry is worth 4% - 7% of the Leaving Cert. It appears on Paper 2. Geometry 1. Definitions You must learn 6 definitions (along with their examples) off by heart. 1) Theorem: A theorem is a rule that has been proved following a certain number of logical steps or by using a previous theorem or axiom that you already know. eg. The angles in a triangle add up to 180 2) Axiom: An axiom is a rule or statement accepted without any proof. eg. There are 360 in a full circle. 3) Corollary A corollary is a statement that follows from a previous theorem. eg. One theorem states that in a parallelogram, opposite sides are equal and opposite angles are equal. A corollary of this is that a diagonal divides a parallelogram into two congruent triangles. 4) Converse The converse of a theorem is the reverse of a theorem. e.g. Theorem: If there are two equal angles in a triangle, the triangle is isosceles. Converse: If a triangle is isosceles, there are two equal angles in the triangle. Sometimes a converse isn t true. e.g. Theorem: In a square, opposite sides are equal. Converse: If opposite sides are equal, then it is a square. This is not true, as it could be a rectangle! 5) Proof A proof is a series of logical steps we use to prove a theorem. e.g. No example needed. The Dublin School of Grinds Page 197

200 6) Implies To imply something is to use something we have proved previously. The symbol for implies is => or e.g. a + b = 15 3 => a + b = 5 The Dublin School of Grinds Page 198

201 2. Constructions At Leaving Cert Ordinary Level there are 21 constructions. These must be learned off by heart. The best way to do this is by looking at a video of the constructions being done. These are all available on The Dublin School of Grinds website: The Dublin School of Grinds Page 199

202 3. Theorems There are 11 theorems. You don t need to learn off by heart, but you do need to understand them, as you can be asked general questions about them. 1) The largest angle is across from the largest side & The smallest angle is across from the smallest side This is best illustrated with a diagram, as follows: Example 1 Which side is the largest in the following diagram: Answer: [xy] 2) The sum of any two lengths of a triangle must be greater than the other side. This is best illustrated with a diagram, as follows: Example 2 The sides of a triangle are 5, 10 and x where x N a) What is the largest possible value for x must be > x => 15 > x or you could say x < 15 The question said x N => x = 14 b) What is the smallest possible value for x 5 + x must be > 10 => x > 5 If x N =>x = 6 c) What is the possible range of values for x? We know from (a) and (b) above that : x 6 and x > > > 8 The Dublin School of Grinds Page 200

203 3) If a line is cut in two equal parts by three parallel lines, then any other line that those parallel lines cross will be cut into equal parts also: This is best explained with an example: Question 3.1 x, y & z are parallel lines. 1, 2 & 3 are three transversals intersecting x, y & z. AB = BC, DE = 7cm and GH = 6cm Find a) EF b) DG The Dublin School of Grinds Page 201

204 Question 3.2 In the diagram, x, y & z are parallel lines. They make intercepts of the indicated lengths on the lines A & B. PQ is parallel to A. a) Write down the length of [PQ] b) Write down the length of [PR] 4) If the side of a triangle is cut by a line parallel to the edge of the triangle in a ratio a:b, then the other side is also cut in the ratio a:b Example 3 This rule leads to many other rules. We will use the figures from the example above to show these: top length top length = bottom length bottom length 3 1 = 6 2 ie. 3 = 3 TRUE bottom length top length 1 3 = 2 6 = bottom length top length ie. 1 3 = 1 3 TRUE The Dublin School of Grinds Page 202

205 bottom length overall length 1 4 = 2 8 = bottom length overall length ie. 1 4 = 1 4 TRUE overall length bottom length 4 1 = 8 2 ie. 4 = 4 TRUE top length overall length = = overall length bottom length top length overall length 3 4 = 6 8 ie. 3 4 = 3 4 TRUE overall length top length 4 3 = = 4 3 TRUE = overall length top length Again we must be able to apply these rules to problems: Example 4 Find x correct to 2 decimal places let s use top length bottom length = top length bottom length = x => 3(6) = 3.8(x + 3) 18 = 3.8x = 3.8x 6.6 = 3.8x = x x = 1.74 The Dublin School of Grinds Page 203

206 Question 3.3 Find the length of the line segments labelled x in each triangle, leave answer in fraction form if required. a) b) c) The Dublin School of Grinds Page 204

207 Question 3.4 Find two possible values of a in the triangle below The Dublin School of Grinds Page 205

208 5) Similar Triangles Similar triangles are triangles that have equal corresponding angles. The corresponding sides are in the same ratio. Similar triangles are also known as equiangular triangles. This is best illustrated with an example. From this rule we can use many ratios: Example 5 small base small right = big base big right 6 12 = 5 10 ie. 1 = Example 6 small left = big left 8 16 = 5 10 ie. 1 = small right big right We could also do ratios within the same triangles: Example 7. small left small bottom = big left big bottom 8 6 = ie. 4 = As you can see there are many ratios we can use. Earlier we saw parallel lines cutting the sides of triangles in the same ratio. This also means that the triangle was divided into two similar triangles: ie: Since the angles are the same, and are similar triangles. The Dublin School of Grinds Page 206

209 Question 3.5 Triangles ABC and DEF are similar a) Find DF b) Find ED The Dublin School of Grinds Page 207

210 Question 3.6 A research team wish to calculate the height of a mountain as follows: They use a laser light source at point A, mounted on a stand 2 metres high. They shine a beam through the top of a pole at B which is 10 metres away from the laser light source. The height of the pole is 20 metres. The distance between the pole and the mountain centre is known from a map to be 1km. Find the height of the mountain h. (Note: diagram not to scale) A B The Dublin School of Grinds Page 208

211 6) Area of a triangle = 1 base x perpendicular height. 2 It does not matter what base you use, once you use the corresponding perpendicular height. Question.3.9 Space for solution on next page The Dublin School of Grinds Page 209

212 The Dublin School of Grinds Page 210

213 7) A diagonal of a parallelogram bisects the area area of 1 = area of 2 8) The area of a parallelogram is the base by the height Note: This is on page 8 of the log tables. 9) A tangent to a circle makes a right angle with the radius 10) A line going through the centre of a circle, which meets a chord at right angles, will cut the chord in half 11) If two circles intersect at one point, their centres and the point of contact lie in a straight line: External contact Internal contact The Dublin School of Grinds Page 211

214 4. Basic rules of angles and triangles Rule 1) There are 180 in a straight angle. Example 1 Let s say we were asked to find the value of x: 6x x + 17 = x + 40 = x = x = 140 x = x = 14 Rule 2) There are 90 in a right angle. Example 2 Let s say we were asked to find the value of x: x = 90 5x = x = 20 x = 20 5 x = 4 Rule 3) There are 360 in a full circle. Example 3 Let s say we were asked to find the value of x: 5x x x x + 30 = x + 80 = x = x = 280 x = x = 20 The Dublin School of Grinds Page 212

215 Rule 4) Vertically opposite angles are equal. Example 4 A = B C = D Rule 5) Alternate angles (or Z angles) are equal. Example 5: Example6: A = B C = D Note: These rules area only true when lines are parallel as shown by or Rule 6) Corresponding angles (or F angles) are equal. Example 7: Example8: A = B C = D Rule 7) Interior angles (or C angles) add up to 180. Example 9: a b a +b =180 The Dublin School of Grinds Page 213

216 Rule 8) There are 180 in any triangle in the world. Example 10 A + B + C = 180 Rule 9) An equilateral triangle (also known as an equiangular triangle) is a triangle where all angles are equal (60 ). Example 11 A = 60 B = 60 C = 60 Note: equal sides are often labelled with little dashes as shown in the diagram above. Sometimes they use two dashes Rule 10) An isosceles triangle is a triangle which has two equal sides. This means that there are two equal angles also. Example 12 Example 13 A = C B = C Rule 11) The exterior angle of a triangle equals the two opposite interior angles added together. Example 14 A = C + D The Dublin School of Grinds Page 214

217 Rule 12) Right angled triangles One of the angles is 90 and the other two angles add up to 90 b a + b = 90 a Question 5.1 Find a and b in the diagram below: Question 5.2 Find x and y in the diagram below: Question. 5.3 Find p and q in the diagram below: The Dublin School of Grinds Page 215

218 5. Quadrilaterals Quadrilaterals have 4 sides. Quadrilaterals have 4 vertices (which is a fancy word for corner points). Quadrilaterals have two diagonals. The 4 angles in a quadrilateral add up to 360. You need to know rules about special quadrilaterals: Type 1: A Square All sides are equal in length Opposite sides are parallel All angles are 90 Diagonals bisect each other (bisect means cuts exactly in half): 90 angles are formed Type 2: A Rectangle Opposite sides are equal in length Opposite sides are parallel All angles are 90 Diagonals bisect each other (bisect means cuts exactly in half) Type 3: A Parallelogram (this is a pushed over rectangle) Opposite sides are equal in length Opposite sides are parallel Opposite angles are equal Diagonals bisect each other (bisect means cuts exactly in half) The angles beside each other add to 180 A + B = 180 or B + C = 180 or D + C = 180 or A + D = 180 The Dublin School of Grinds Page 216

219 Type 4: A Rhombus (this is a pushed over square) All sides are equal in length Opposite sides are parallel Opposite angles are equal Diagonals bisect each other (bisect means cuts exactly in half): 90 angles formed. The angles beside each other add to 180 A + B = 180 or B + C = 180 or D + C = 180 or A + D = 180 Question 6.1 Find the value of y in the diagram below: Question 6.2 Find the angles a, b and c in the diagram below: The Dublin School of Grinds Page 217

220 Question 6.3 Find the value of p and q in the diagram below: The Dublin School of Grinds Page 218

221 6. Basic rules of circles Again, we should recall some information about circles from the Junior Cert. You must know the following points: i) Radius radius ii) Diameter diameter iii) Chord chord iv) Circumference circumference v) Arc arc vi) Tangent tangent vii) Sector sector viii) When standing on the same arc, an angle at the centre of a circle is twice an angle at the circumference. x 2x The Dublin School of Grinds Page 219

222 ix) When standing on the same arc, an angle at the circumference is equal to any other angle at the circumference. x x x) An angle opposite a diameter in a circle is 90. O o xi) Opposite angles of a cyclic quadrilateral (also known as a four sided shape which touches the circle at its corner points) in a circle add up to 180 : A D B A + C = 180 B + D = 180 C xii) A tangent to a circle makes a right angle with the radius. The Dublin School of Grinds Page 220

223 7. Transformations There are 3 types of transformations: i) Translations ii) Axial Symmetry iii) Central Symmetry i) Translations A translation moves a figure in a straight line. It looks the same as before: ii) Axial symmetry Axial symmetry reflects a figure through a line. It looks back to front: If you draw a line through a figure and it is balanced on either side of the line then you have an axis of symmetry. eg. B A square has 4 axis of symmetry The letter B has 1 axis of symmetry The letter f has no axis of symmetry The Dublin School of Grinds Page 221

224 Question 9.1 Draw a shape which has exactly 3 axes of symmetry. Show the axes on the diagram. Question 9.2 Draw a shape which has exactly 2 axes of symmetry. Show the axes on the diagram. The Dublin School of Grinds Page 222

225 iii) Central Symmetry Central symmetry reflects a figure through a point. It looks back to front and up side down. If you draw a point in a figure and the figure is balanced around it, then you have the centre of symmetry. The Dublin School of Grinds Page 223

226 8. Enlargements What s enlargement? Enlargement, sometimes called scaling, is a kind of transformation that changes the size of an object. The image created is similar* to the object. Despite the name enlargement, it includes making objects smaller. *Shapes are similar if their corresponding angles are equal. Their corresponding sides are then in the same ratio. For every enlargement, a scale factor must be specified. The scale factor is how many times larger than the object the image is. For any enlargement, there must be a point called the centre of enlargement. The centre of enlargement can be anywhere, but it has to exist. If you are asked to enlarge an object in an exam, the question will always tell you which centre of enlargement to use. The Dublin School of Grinds Page 224

227 How do I actually find and draw the enlarged image? With a ruler, draw a straight line from the centre of enlargement to each point on the object. The Dublin School of Grinds Page 225

228 Important rules: Scale factor (k) = Image length Object length Image area = k 2 Object area (New area =old area scale factor 2 ) The Dublin School of Grinds Page 226

229 9. Past and probable exam questions Question 1 The Dublin School of Grinds Page 227

230 Question 2 The Dublin School of Grinds Page 228

231 Question 3 The Dublin School of Grinds Page 229

232 Question 4 The Dublin School of Grinds Page 230

233 Question 5 The Dublin School of Grinds Page 231

234 The Dublin School of Grinds Page 232

235 Question 6 The Dublin School of Grinds Page 233

236 Question 7 The Dublin School of Grinds Page 234

237 Question 8 The Dublin School of Grinds Page 235

238 Question 9 The Dublin School of Grinds Page 236

239 Question 10 The Dublin School of Grinds Page 237

240 Question 11 The Dublin School of Grinds Page 238

241 Question 12 The Dublin School of Grinds Page 239