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

Transcription

1 6 th Year Maths Higher 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-h-Strand 2 of 5).

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 Your Leaving Cert paper is based on the syllabus published by a gang called the NCCA. The NCCA have the syllabus broken into 5 sections, which are called the 5 strands. Strand 2 is worth 23% to 46% of The Leaving Cert. It appears on Paper 2. Contents Trigonometry 1) Pythagoras ) Sin, Cos, Tan ) Radians ) Angles of greater than 360 (or 2π radians) ) Trigonometric Equations ) The Sine Rule ) The Cosine Rule ) Area of a triangle ) Sector area/arc length ) 3-D problems ) Trigonometric function graphs ) Derivation of Trigonometry Formulae ) Application of formulae ) Past and probable exam questions ) Solutions to Trigonometry Co-ordinate Geometry of the Line 1) Distance between two points ) Midpoint of a line ) Slope of a line ) Slopes of parallel lines ) Slopes of perpendicular lines ) The equation of a line ) The point of intersection of a line and the axes ) Point of intersection of two lines ) The area of a triangle ) Perpendicular distance from a point to a line ) The angle between two lines ) Dividing a line segment in a ratio ) 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) and radius r ) Equation of a circle with centre (h, k) and radius r ) Equation in the form x 2 + y 2 + 2gx + 2fy + c = ) Tangent to a circle ) Point(s) of intersection of a circle with an axis ) Point(s) of intersection of a circle and a line ) Points inside/outside/on a circle ) Locus ) The equation of a tangent ) Touching Circles ) Problems in g, f and c ) Past and probable exam questions ) Solutions to Co-ordinate Geometry of the Circle Geometry 1) Theorems ) Definitions ) Constructions ) Basic rules of angles and triangles ) Congruent triangles ) Quadrilaterals ) Parallel line rules ) Similar triangles ) Area of triangles and parallelograms ) Basic rules of circles ) Transformations ) Enlargements ) Past and probable exam questions ) Solutions to Geometry The Dublin School of Grinds Page 2

5 Trigonometry Trigonometry is worth 7% to 18% of The Leaving Cert. It appears on Paper 2. 1) Pythagoras On page 16 of the Formulae and Tables Booklet you will see the following: c 2 = a 2 + b 2 (longest side) 2 = (other side) 2 + (other side) 2 Example 1: A kite is held by a string 13m long. When blown in the wind it is 9m above the ground as shown in the diagram: How far is the point A from B, correct to the nearest metre? List any assumptions you have made. Solution: 13 2 = x = 81 + x = x 2 88 = x 2 88 = x = x x = 9 meters Assumptions: I assumed the string is straight. I assumed the ground is level. The Dublin School of Grinds Page 3

6 2) Sin, Cos, Tan On page 16 of The Formulae and Tables Booklet you will see the following: What this really means is: The way to remember this is: Sin A = Cos A = opposite hypotenuse adjacent hypotenuse Tan A = opposite adjacent SOH CAH TOA or Silly Old Harry, Caught A Herring, Trawling Off America or Some Old Horses, Can Always Hear, Their Owners Approaching Labelling the sides: i.e. The hyp is the side across from the right angle. The opp is the side across from the angle in question. The adj is the other side. Note: Don t use the letter h for hypotenuse as you may get it mixed up with h for height when you re under pressure in the exam. The Dublin School of Grinds Page 4

7 Example 1: Find the measure of the angle x. Give your answer correct to two decimal places. Solution: The sides in question are hyp and opp, so we will use Sin: Sin x = opp hyp Sin x = 4 15 x = Sin 1 ( 4 15 ) = Note: You must be able to change angles from decimals to a thing called minutes. Example 2: Change to the nearest minute. Casio Calculator: Sharp Calculator: To change from decimals to minutes, type in the number with the decimal, then press the button with all the commas (2 buttons above number 8), then press equals. To change from decimals to minutes, type in the number with the decimal, then press 2 nd function, then press DMS. You will get " Now the 37 is the degrees, the 21 is the minutes and the 36 is a thing called seconds. If your value for seconds is 30 or greater you round up. If it is less than 30 you round down. Therefore we will round up: = Question 2.1 Change to the nearest minute. The Dublin School of Grinds Page 5

8 Question 2.2 Change to the nearest minute. Sometimes an angle will be given to you in degrees and minutes so you have to be careful when putting this into your calculator. Example 3: Find Sin31 26 correct to 4 decimal places. Casio Calculator: Press Sin, then 31, then the comma button, then 26, then the comma button again, then equals. Sharp Calculator: Press Sin, then 31, then the DMS button, then 26, then the DMS button again. You will get (depending on your calculator you may even get more decimal places) We were asked to round off to 4 decimal places so our answer is Question 2.3 Evaluate sin35 7 to four decimal places. There are three other points we must know to help us with the real life scenarios which are at the heart of Project Maths: i) Angle of elevation: This is the angle your line of vision makes with the horizontal when looking up: ii) Angle of depression: This is the angle your line of vision makes with the horizontal when looking down: The Dublin School of Grinds Page 6

9 iii) Compass directions: For example: Example 4: A boat is anchored off the coast near100m high cliffs in Co. Kerry. The angle of elevation from the boat to the cliff top is 15. The next day the boat moves further towards land and anchors again. This time the angle of elevation is 38. How far did the boat move? Give your answer correct to the nearest metre and make assumptions as necessary, stating them clearly. Solution: Take out the triangle on the right: Tan 38 = 100 x 100 x = Tan 38 x = m The Dublin School of Grinds Page 7

10 Now take out the overall triangle: Tan 15 = y y = Tan 15 y = m The distance the boat moves is y x: = = metres 245 metres Assumptions: It was assumed the cliffs were vertical, thus giving a right angle where they meet at the sea. Question 2.4 The beach at Brittas Bay slopes downwards at a constant angle of 12 to the horizontal. How far out horizontally into the sea can a man walk before the water covers his head? Make assumptions as necessary. The Dublin School of Grinds Page 8

11 3) Radians Just as distance can be measured in miles or kilometers, (1km miles) angles can be measured in degrees or radians. (1 radian 57 ) You must remember: 1 = π radians π radians = Note: This π isn t equal to 3.14 like it is when we are dealing with circles, so don t get them mixed up!! Example 1: Example 2: Convert 40 to radians Convert 5π radians to degrees 9 1 = π radians 180 ) radians π radians = 180 => 40 = 40( π 180 = 40π radians => 5π 180 radians = 100 = 2π 9 9 radians = 5 9 (180 ) Question 3.1 (i) Convert 60 to radians (ii) Convert π radians to degrees 4 (iii) How many radians are there in a full rotation? The Dublin School of Grinds Page 9

12 4) Angles of greater than 360 (or 2π radians) In trigonometry we have four quadrants: For example: 370 would do a full rotation plus another 10. It terminates (or stops) in Quadrant would terminate in Quadrant 3-60 would terminate in Quadrant would terminate in Quadrant 3 6π 5 radians would terminate in Quadrant 3 Question 4.1 In which quadrant would the following angles terminate? (i) -880 (ii) (iii) 14π 3 17π 7 radians radians The Dublin School of Grinds Page 10

13 5) Trigonometric Equations Consider the following diagram (Sex And The City): A + stands for All plus S + stands for Sin plus T + stands for Tan plus C + stands for Cos plus To read angles in any quadrant we use the following diagram: To solve trigonometric equations we will use these 2 diagrams. These equations are best explained with some examples. Example 1: Solve for θ: Cos θ = 1, 0 θ Step 1: Find the reference angle by finding the inverse of the value given: => Reference angle = Cos 1 ( 1 2 ) = 45 Step 2: Use the SATC diagram to establish where Cos is positive: Step 3: Use the reference diagram: θ = or θ = =45 = 315 The Dublin School of Grinds Page 11

14 Question 5.1 Solve for θ: sinθ = 3, 0 θ Next let s try one with a minus sign: Example 2: Solve for θ: Sin θ = θ 360 Step 1: Find the reference angle by finding the inverse of the value given and by ignoring the minus sign. Reference angle = Sin 1 ( 1 ) 2 = 45 Step 2: Use the SATC diagram to establish where sin is negative: Step 3: Use the reference diagram or Question 5.2 Solve for θ: cosθ = 1, 0 θ The Dublin School of Grinds Page 12

15 Question 5.3 Find correct to one decimal place, two values of A for CosA = 3, where 0 A There may also be a bit of rearranging to be done before starting the question. Let s do an example to show this Question 5.4 Solve, correct to one decimal place: 7Cos θ + 3 = 0 where 0 θ 360 The question can also be given in radians rather than degrees. You should do the questions in degrees, because radians are a load of crap. Then just change to radians at the end to get the full marks. The Dublin School of Grinds Page 13

16 Question 5.5 Solve for θ: Tan θ = 1, 0 θ 2π 3 Question 5.6 Solve for θ: (θ in radians) if Cos θ= 1, 0 θ 2π 2 The Dublin School of Grinds Page 14

17 The Examiner can put numbers in front of θ, as in our next example: Example 3 Solve for θ: if Sin 2θ = 1, 0 θ Step 1: Start as normal, by finding the inverse of the given value and by ignoring the 2 in front of the θ: => reference angle =Sin 1 ( 1 2 ) Step 2: Again, this step is as normal: = 30 Step 3: This step is where the 2θ comes into play: Step 3: 2θ = or 2θ = = 30 = 150 Don t divide by the 2 yet because this is special! Therefore we need a special step: Step 4: Given 0 θ 360 Multiply by 2: 0 2θ 720 This means our solutions for 2θ must be between 0 and 720 We know the angles repeat themselves every 360 2θ = 30 2θ = 150 or or 2θ = θ = = 390 = 510 or or 2θ = θ = = 750 = 870 Too high!! Too high!! 2θ = 30, 150, 390, 510 Now we can divide, by the 2: θ = 15, 75, 195, 255 Question 5.7 Solve for θ: cos3θ = 1, 0 θ The Dublin School of Grinds Page 15

18 Question 5.8 Solve for θ: cos θ = 1, 0 θ Sometimes the Examiner doesn t include a range for θ. i.e. He won t say 0 θ 360 or 0 θ 2π. Therefore we must give a general solution. Let me explain Let s say we find the answer to a question to be 30 and 150. We could also add 360 onto these an infinite number of times. Therefore the general solution would be: Where n ϵ Z θ = n and θ = n i.e. we add on 360 n times (since 360 is a full rotation in degrees). If the question was in radians we would add 2πn (since 2π is a full rotation in radians). We write n ϵ Z because there could be positive or negative angles. Question 5.9 Solve for θ: cosθ = 1 2. The Dublin School of Grinds Page 16

19 The Examiner can also ask you to solve equations involving functions such as Sec and Cot. Example 4: Solve Sec θ = 2 0 θ 360 From page 13 of the Formulae and Tables booklet: 1 Sec A = Cos A 1 Cos θ = 2 1 = 2 Cos θ Cos θ = 1 2 We can then continue as before Question 5.10 Find the two values of x correct to two decimal places if 0 x 360 (a) 17 Cot x = -22 (b) 3 Sec x 5 = 0 And how about this beauty to mix it all together: Question 5.11 Solve for θ, 2cot 2 θ = 1, correct to 4 decimal places, where θ is in radians. The Dublin School of Grinds Page 17

20 The Examiner can give you one of these equations in a hidden way, by using a diagram: Question 5.12 The diagram shows the graph of the function g(θ) = sin 2 θ sinθ. Calculate the value of M and the value of N. Give your answer in degrees. The Dublin School of Grinds Page 18

21 Question 5.13 The diagram below shows the graph of the function f(θ) = cosθsin2θ. Calculate the values of P, Q, R, S and T. Give your answers in degrees. The Dublin School of Grinds Page 19

22 6) The Sine Rule On page 16 of the Formulae and Tables booklet you will see the following: a SinA = b SinB = c SinC From correcting exams, I see that this can lead to confusion amongst some students, since questions in the Leaving Cert often include the letters a, b, c, A, B, C, when they have nothing to do with the above diagram and formula! The way to remember the Sine Rule is as follows: any side the sin of the angle across from it = any side the sin of the angle across from it Let s illustrate this with an example. Example 1: Find x, correct to three decimal places. Using the Sine rule: x Sin30 = 7 Sin110 x = 7Sin30 Sin110 = 3.725cm The Dublin School of Grinds Page 20

23 Question 6.1 In order to calculate the height of the lighthouse in the diagram below, the measures of the angles of elevation of the top of the lighthouse from points A and B on level ground were taken. AB = 15meters What equipment is used to take such measurements? Find the height of the building, h, in terms of α and β. The Dublin School of Grinds Page 21

24 Question 6.2 Two surveyors want to find the height of an electricity pylon. There is a fence around the pylon that they cannot cross for safety reasons. The ground is inclined at an angle. They have a clinometer (for measuring angles of elevation) and a 100-meter tape measure. They have already used the clinometer to determine that the ground is inclined at 10 to the horizontal. i) Explain how they could find the height of the pylon. Your answer should be illustrated on the diagram. Show the points where you think they should take the measurements, write down what measurements they should take and briefly outline how these can be used to find the height of the pylon. ii) Write down possible values for the measurements taken and use them to show how to find the height of the pylon. (That is, find the height of the pylon using your measurements and showing your work.) The Dublin School of Grinds Page 22

25 The ambiguous case When we are given one angle and two sides, the Sine Rule may give an incorrect answer. This is called the ambiguous case. This situation arises when we are given two sides and a non-included angle. Let s look at a simple example. In a ABC, AB = 82, BC = 65 and BAC = 50 there are two possible positions for vertex C, namely C 1 and C In the triangle above, AC 1 B = 75.1 and AC 2 B = Question 6.3 = It s the final game of The Heineken Cup rugby tournament and Leinster have been given a penalty kick in the last minute. If they score, the cup is theirs. All week during training, Johnny Sexton has been measuring how far he can kick for the posts. He is confident he has a range of 45 metres. The diagram below shows the details from Sky Sports: Based on this information, is there a chance Sexton can score the penalty? Justify your answer. The Dublin School of Grinds Page 23

26 7) The Cosine Rule On page 16 of the Formulae and Tables booklet you will see the following: In words this means: a 2 = b 2 + c 2 2bcCosA (any side) 2 = (other side) 2 + (other side) 2 2(other side)(other side)cos(angle across from the first side you picked) Let s illustrate this with an example Example 1: Find x, correct to 2 decimal places: Solution: Using the Cosine rule: x 2 = (5) 2 + (9) 2 2(5)(9)Cos70 x 2 = Cos70 x 2 = x = x 8.67cm Question 7.1 Find the measure of the angle θ: The Dublin School of Grinds Page 24

27 Question 7.2 An engineer is designing a bridge over a river. She needs to know the distance between two locations, Y and Z, which are on the other side of the river (see diagram below). She finds two locations, W and X, on her side of the river which are on the same level such that WX = 20 meters. Using a theodolite, she finds the measure of four angles as follows: YWZ = 48 XWZ = 57 WXY = 63 YXZ = 50 Use this information to find YZ, showing all your work clearly. Y Z W 57 20m 63 X The Dublin School of Grinds Page 25

28 Question 7.3 Note: More space for your solution on the next page. The Dublin School of Grinds Page 26

29 The Dublin School of Grinds Page 27

30 8) Area of a triangle The basic formula for the area of a triangle is: area of triangle = 1 (base)(perpendicular height) 2 However, sometimes we aren t given the perpendicular height. Therefore another formula we can use is: area of triangle = 1 2 absinc This is given on page 16 of the Formulae and Tables booklet. I prefer to remember it in words as follows: area of triangle = 1 (one side)(another side)(sin of the angle in between these two sides) 2 Let s illustrate this with an example. Example 1: Find the area of the following triangle, correct to the nearest cm cm 7cm area = 1 2 (10)(7)Sin70 33cm 2 The Dublin School of Grinds Page 28

31 9) Sector area/arc length The following formulae are given on page 9 of the Formulae and Tables booklet: Area of sector Length of arc When θ is in degrees A = πr 2 ( θ 360 ) L = 2πr ( θ 360 ) When θ is in radians A = 1 2 r2 θ L = rθ Example 1: Find the area of a sector of 30 of a circle of radius 9cm, correct to one decimal place. A = πr 2 ( θ 360 ) = π(9) 2 ( ) = 21.2cm 2 Question 9.1 The circle in the diagram below has radius 17cm. The two sectors have arc length l 1 and l 2 as shown. The areas of the sectors are equal. The angle given is in radians. Find i) θ in radians ii) l 1 : l 2 The Dublin School of Grinds Page 29

32 Question 9.2 When bored, the Dublin Port Tunnel can be assumed to have been cylindrical in shape. To make the road surface, an engineer needs to calculate the amount of gravel needed. His architect gives him the following sketch and information: Diameter of tunnel = 6m Length of tunnel = 4.5km Gravel 0.7 metres Estimate, in m 3, the volume of gravel needed. The Dublin School of Grinds Page 30

33 Question 9.3 The diagram below shows the outline of the shot-put field being used for the Rio Olympics. The lines BC and AC can be assumed to be tangents to the circle. (a) Find the area of the throwing area in terms of θ, where θ is in radians (b) Find the area of OBC, in terms of Tan θ (c) Show that the landing area can be represented by 9(Tan θ θ)m 2 C Landing area B Throwing area O 2θ A The Dublin School of Grinds Page 31

34 10) 3-D problems These questions can be very handy once we can identify where the right angles are. From there we will use the rules we have learnt so far (Pythagoras, Sin, Cos, Tan, Sine rule, Cosine rule). Example 1: Below is the outline of a rectangular box: A AB = 5cm BF = 2cm FG = 10cm D E B H C a) Find AF correct to one decimal place b) Find AG correct to one decimal place c) Find AGE correct to the nearest degree d) Find BGF correct to the nearest degree F G Solution: a) Look at ABF Using Pythagoras: x 2 = x 2 = x 2 = 29 x = 29 x = AF 5.4cm b) Look at ACG CG = 2cm since the height is consistent on all sides of the box: I know that if I can find AC then I will be able to find AG. So now I will look at ABC BC = 10cm since the length is constant on all sides of the box: The Dublin School of Grinds Page 32

35 Using Pythagoras: y 2 = y 2 = y 2 = 125 y = 125 y = Note: Don t round off yet, since this is not an answer Now look back at ΔACG Using Pythagoras: z 2 = z 2 = 129 z = 129 z = AG 11.4cm c) We can use Sin, Cos or Tan here. Let s use Sin: SinW = opp d) Look at BGF SinW = hyp 2 W = Sin 1 ( ) 11.4 W = AGE 10 Since we know the opp and the adj let s use Tan: TanV = opp adj TanV = 2 10 V = Tan 1 ( 2 10 ) V = BGF 11 The Dublin School of Grinds Page 33

36 Question 10.1 The diagram shows a pyramid. The base, ABCD, is a horizontal square of side 10cm. The vertex V is vertically above the midpoint, M, of the base. VM = 12cm. Calculate the size of VAM, correct to the nearest degree. The Dublin School of Grinds Page 34

37 Question 10.2 The diagram represents a prism. AEFD is a rectangle. ABCD is a square. EB and FC are perpendicular to the plane ABCD. AB = AD = 60cm. ABE = 90. BAE = 30. Calculate the size of the angle that the line DE makes with the plane ABCD. The Dublin School of Grinds Page 35

38 Question 10.3 The diagram represents a cuboid ABCDEFGH. CD =5cm, BC =7cm, BF =3cm. a) Calculate the length of AG. Give your answer correct to 2 decimal places. b) Calculate the size of the angle between AG and the face ABCD. Give your answer correct to 2 decimal places. The Dublin School of Grinds Page 36

39 Question 10.4 The diagram shows a door-wedge with a rectangular horizontal base PQRS. The sloping face PQTU is also rectangular. PQ = 3.8cm and TQR = 6. The height TR is 2.5cm. Calculate the length of the diagonal PT. The Dublin School of Grinds Page 37

40 11) Trigonometric function graphs The Examiner can ask you to graph trigonometric functions. Example 1: Graph the function y=sinx, 0 X 360. We sub in numerous values for X and calculate the corresponding values for y using our calculator: X y=sinx In this example I have gone to two decimal places. We may be told to round off or even go to more decimal places. We can now graph this: We could also be asked to use a range in radians (0 X 2π) rather than degrees (0 X 360 ). Just make sure your calculator is in radian mode! Next, let s graph the cos function (in radians): Question 11.1 Sketch the function y=cosx, 0 X 2π π π X π 4 y=cosx π 5π 4 3π 2 7π 4 2π The Dublin School of Grinds Page 38

41 As you will see, the graphs of Sin and Cos repeat themselves over regular intervals, therefore they are called periodic functions. The period of both SinX and CosX is 360 (or 2π radians). You will also see they have a maximum and a minimum height. This is called the range. The range of both SinX and CosX is [-1, 1]. Note 1: The amplitude is the distance from the midway line to the top or bottom. Note 2: If there is a minus in front the function (eg: -sinx), then the graph would be inverted, ie: turned inside out. (ie: SinX starts uphill, -SinX starts downhill. CosX starts downhill, -CosX starts uphill). Note 3: At the y-axis, Sine graphs will be at the midway line. At the y-axis, Cosine graphs will be at the maximum or minimum. The Examiner can also ask you to graph functions in the form ncosx (e.g. 2CosX) or CosnX (e.g. Cos2X). Let s look at two examples to see how this affects the graphs: Example 2 Graph the function 2CosX where 0 X 360. X y=2cosx As you can see, the period is still 360. But the range is now [-2, 2]. Example 3 Graph Cos2X where 0 X 360 : X y=cos2x As you can see, the period is now 180. The range is still [-1, 1]. The Dublin School of Grinds Page 39

42 Question 11.2 The function f(x) = 3Sin(2X) is defined for x ϵ R Complete the table below: X 0 2X π 4 π 2 3π 4 π Sin(2X) 3Sin(2X) Draw the graph of y=f(x) in the domain 0 X π, X ϵ R Write down the range and period of f. Range = Period = The Examiner can also ask you to graph functions in the form n + cosx (eg: 2 + cosx) let s try an example to see how this affects the graphs: Question 11.3 Graph the function: 2 + sinx for 0 x 360. So to summarise: For a graph in the form a + b sin cx or a + b cos cx, the value of a gives the vertical translation (which is the distance from the x-axis to the midway line), the value of b gives the amplitude (which is the distance from the top or bottom to the midway line), and the value of c comes from the period, where period is 360 for radians). The Dublin School of Grinds Page 40 c (or 2π c

43 Question 11.4 A trigonometric function is shown, where x is in degrees. The co-ordinates of the points A and B are (-60,3) and (30,11) respectively. (i) State whether f(x) is a sine function or a cosine function. Give a reason. (ii) Write the equation of the graph. (iii) Find the co-ordinates of the points C and D. The Dublin School of Grinds Page 41

44 The Examiner can relate a function to a real life scenario, as seen in the next example: Question 11.5 In a particular area of woodland, the population of woodmice and owls was monitored every month for two years. It is know that woodmice represent the major prey for owls in the area. The results are shown on the graphs in the diagram below. The Dublin School of Grinds Page 42

45 The Examiner can mix these graphs with trigonometric equations (from section 5): Question 11.6 The diagram below shows the graph of the function f: x Sin2X. The line 2y=1 is also shown Find the co-ordinates of the point P in the diagram. The Dublin School of Grinds Page 43

46 We have seen that the graphs of Sin and Cos have similar properties. But what about Tan? Example 4: Graph the function y = TanX, 0 X 360. X y=tanx x x As you will see, Tan90 and Tan270 are undefined on your calculator. This means the graph never reaches those points. The dotted lines that the curve never touches are called asymptotes. As you can see, the period is 180. There are no maximum or minimum values, therefore there is no range. Note: We will not be asked to work with functions in the form ntanx (e.g. 2TanX) or TannX (e.g. Tan2X) or n+tanx (e.g. 2 + Tanx). The Dublin School of Grinds Page 44

47 12) Derivation of Trigonometry Formulae There are 8 formulae the Examiner can ask you to derive. Below you will see simplified, full mark versions of these. Make sure to include the diagrams. You do not have to include the parts labelled as Notes. 1) To prove Cos 2 A + Sin 2 A = 1 From the unit circle: (Cos) 2 A + (Sin) 2 A = 1 2 Cos 2 A + Sin 2 A = 1 2) To prove The Sine Rule: Case 1: A is acute a SinA = b SinB = c SinC Case 2: A is obtuse b y a y b a A B E A B SinA = y b bsina = y SinB = y a asinb = y asinb = bsina a SinA = b SinB E = 180 A SinE = Sin(180 A) = SinA Note: The line above actually comes from working with the compound angle formulae on page 14 of the Formulae and Tables Booklet but you don t need to show the workings. SinA = y b bsina = y SinB = y a asinb = y asinb = bsina a SinA = b SinB The Dublin School of Grinds Page 45

48 3) To prove: The Cosine Rule: a 2 = b 2 + c 2 2bcCosA Note: Place the triangle with angle A placed at the origin and side c placed along the X-axis. Draw a circle of radius b around the origin. As this circle is b times larger than the unit circle, the x and y coordinates will also be b times larger. The coordinates of the triangle are (0, 0), (c, 0) and (bcosa, bsina). distance = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 O A b (bcosa, bsina) c a (c, 0) a = (bcosa c) 2 + (bsina 0) 2 a 2 = b 2 Cos 2 A 2bcCosA + c 2 + b 2 Sin 2 A = b 2 (Cos 2 A + Sin 2 A) + c 2 2bcCosA = b 2 (1) + c 2 2bcCosA a 2 = b 2 + c 2 2bcCosA 4) To Prove: Cos(A B) = CosA CosB + SinA SinB Using the distance formula: P(CosA, SinA) Q(CosB, SinB) distance = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 PQ 2 = (CosA CosB) 2 + (SinA SinB) 2 = Cos 2 A 2CosA CosB + Cos 2 B + Sin 2 A 2SinA SinB + Sin 2 B = (Cos 2 A + Sin 2 A) + (Cos 2 B + Sin 2 B) 2(CosA CosB + SinA SinB) 1 A-B B A O 1 = 2 2(CosA CosB + SinA SinB) (1) Using the Cosine formula on ΔOPQ: PQ 2 = (1)(1)Cos(A B) = 2 2Cos(A B) (2) Equating (1) and (2): 2 2Cos(A B) = 2 2(CosA CosB + SinA SinB) Cos(A B) = CosA CosB + SinA SinB 5) To prove: Cos(A + B) = CosA CosB SinA SinB Well: Cos(A B) = CosA CosB + SinA SinB Replace B with B: Cos(A ( B)) = CosA Cos( B) + SinA Sin( B) Note: From page 13 of the Formulae and Tables booklet Cos( A) = CosA and Sin( A) = SinA Cos(A + B) = CosA CosB + SinA ( SinB) Cos(A + B) = CosA CosB SinA SinB The Dublin School of Grinds Page 46

49 6) To Prove: Cos2A = Cos 2 A Sin 2 A Well: Cos(A + B) = CosA CosB SinA SinB Replace B with A: Cos(A + A) = CosA CosA SinA SinA Cos2A = Cos 2 A Sin 2 A 7) To Prove: Sin(A + B) = SinA CosB + CosA SinB Well: Cos(A B) = CosA CosB + SinA SinB Replace A with 90 A: Cos(90 A B) = Cos(90 A)CosB + Sin(90 A)SinB Note: Cos(90 A) is always SinA. Similarly Sin(90 A) is always CosA. Cos(90 (A + B)) = SinA CosB + CosA SinB Sin(A + B) = SinA CosB + CosA SinB 8) To prove: Tan(A + B) = TanA+TanB 1 TanA TanB Tan(A + B) = = Sin(A + B) Cos(A + B) SinA CosB + CosA SinB CosA CosB SinA SinB Now just divide everything by CosA CosB. SinA CosB CosA SinB + = CosA CosB CosA CosB CosA CosB SinA SinB + CosA CosB CosA CosB = TanA + TanB 1 TanA TanB The Dublin School of Grinds Page 47

50 13) Application of formulae These types of questions will require us to make use of the formulae on pages 13, 14 and 15 of the Formulae and Tables booklet. Example 1: Write sin240 in surd form, without the use of your calculator. sin(a + B) = sinacosb + cosasinb sin ( ) = sin180cos60 + cos180sin60 = (0) ( 1 2 ) + ( 1)( 3 2 ) = 3 2 Question 13.1 Without using a calculator, show that the following is true: Sin10 + Sin80 = 2Cos35 Example 2: Prove that Tan3θ = 3Tanθ Tan3 θ 1 3Tan 2 θ From page 14: TanA + TanB Tan(A + B) = 1 TanA TanB Tan2θ + Tanθ Tan(2θ + θ) = 1 Tan2θ Tanθ But from page 14: Tan2A = 2TanA 1 Tan 2 A Tan2θ = 2Tanθ 1 Tan 2 θ Tan(2θ + θ) = 2Tanθ 1 Tan 2 θ + Tanθ 1 2Tanθ 1 Tan 2 θ Tanθ Now get rid of the fractions on top and bottom by multiplying top and bottom by 1 tan 2 θ. = 2Tanθ + Tanθ Tan3 θ 1 Tan 2 θ 2Tan 2 θ = 3Tanθ Tan3 θ 1 3Tan 2 Q. E. D. θ The Dublin School of Grinds Page 48

51 Question 13.2 Prove Sin3A+SinA Cos3A+CosA = Tan2A We can also be asked to change products to sums (or differences) and vice versa. Example 3: Express Cos3θ Sin5θ as a sum or difference of two trigonometric functions. From page 15: 2CosA SinB = Sin(A + B) Sin(A B) CosA SinB = 1 2 Sin(A + B) 1 Sin(A B) 2 Cos3θ Sin5θ = 1 2 Sin(3θ + 5θ) 1 Sin(3θ 5θ) 2 = 1 2 Sin(8θ) 1 2 Sin( 2θ) From page 13: Sin( A) = SinA Sin( 2θ) = Sin(2θ) = 1 2 Sin(8θ) Sin(2θ) Question 13.3 Express Cos3xCos2x as a sum. The Dublin School of Grinds Page 49

52 14) Past and probable exam questions Question 1 Question 2 The length of [AC] is 6 metres. And the pitch of the roof is 35, as shown. AD = DE = EC and AF = FB = BG = CG The Dublin School of Grinds Page 50

53 Question 3 Question 4 The Dublin School of Grinds Page 51

54 Question 5 The Dublin School of Grinds Page 52

55 Question 6 The Dublin School of Grinds Page 53

56 Question 7 The Dublin School of Grinds Page 54

57 Question 8 The Dublin School of Grinds Page 55

58 Question 9 The Dublin School of Grinds Page 56

59 Question 10 The Dublin School of Grinds Page 57

60 Question11 The Dublin School of Grinds Page 58

61 Question 12 The Dublin School of Grinds Page 59

62 Question 13 The graphs of three functions are shown in the diagram below. The scales on the axes are not labeled. The three functions are: x Cos3x x 2Cos3X x 3Cos2X a) Identify which function is which, and write your answers in the spaces below the diagram. f: x g: x h: x b) Label the scales on the axes in the diagram in part (a) Question 14 The diagram below shows the graph of the function f: x Cos3X. a) On the same diagram above, sketch the graph of g: x CosX and h: x 2Cos3X. Indicate clearly which is g and which is h. b) Find the coordinates of the point P in the diagram. The Dublin School of Grinds Page 60

63 Question 15 Graph 1 Graph 2 Graph 3 Graph 4 Graph 5 Graph 6 Graph 7 Graph 8 Note: The question is on the next page. The Dublin School of Grinds Page 61

64 Question 16 The Dublin School of Grinds Page 62

65 Question 17 The Dublin School of Grinds Page 63

66 The Dublin School of Grinds Page 64

67 Question 18 Frank, one of the fisher men who works on the boat, is prone to sea sickness. (Interesting career choice ) Frank starts feeling sick if the boat bobs up and down more than 6 times per minute The Dublin School of Grinds Page 65

68 The Dublin School of Grinds Page 66

69 Question 19 Question 20 The Dublin School of Grinds Page 67

70 Question 21 The Dublin School of Grinds Page 68

71 Question 22 Question 23 The Dublin School of Grinds Page 69

72 Question 24 The Dublin School of Grinds Page 70

73 The Dublin School of Grinds Page 71

74 15) Solutions to Trigonometry Question 2.1 = Question 2.2 = Question 2.3 = Question 2.4 ie: tan 12 = 1.8 x x = 1.8 tan 12 x 8.47m Question 3.1 i) ii) iii) 1 = π 180 rad 60 = 60 ( π 180 ) rad = π 3 rad π rad = 180 π 180 rad = 4 4 = 45 1 = π 180 rad 360 = 360 ( π 180 ) rad = 2π rad The Dublin School of Grinds Page 72

75 Question 4.1 i) Quadrant 3 ii) Quadrant 2 iii) Quadrant 4 Question 5.1 Step 1: Reference angle: Step 2: sin 1 ( 3 2 ) = 60 Step 3: θ = = 60 or θ = = 120 Question 5.2 Step 1: Reference angle: Step 2: cos 1 ( 1 2 ) = 45 Step 3: θ = = 135 or θ = = 225 Question 5.3 Step 1: Reference angle: Step 2: cos 1 ( 3 5 ) = 53.1 Step 3: A = = or A = = The Dublin School of Grinds Page 73

76 Question cos θ + 3 = 0 7 cos θ = 3 cos θ = 3 7 Step 1: Reference angle: Step 2: cos 1 ( 3 7 ) = 64.6 Step 3: θ = = or θ = = Question 5.5 Step 1: Reference angle: Step 2: tan 1 ( 1 3 ) = 30 Step 3: θ = = 30 or θ = = = π 180 rad 30 = 30 ( π 180 ) rad = π 6 rad 1 = π 180 rad 210 = 210 ( π 180 ) rad = 7π 6 rad The Dublin School of Grinds Page 74

77 Question 5.6 Step 1: Reference angle: Step 2: cos 1 ( 1 2 ) = 45 Step 3: θ = = 135 or θ = = = π 180 rad 135 = 135 ( π 180 ) rad = 3π 4 rad 1 = π 180 rad 225 = 225 ( π 180 ) rad = 5π 4 rad Question 5.7 Step 1: Reference angle: Step 2: cos 1 ( 1 2 ) = 45 Step 3: Step 4: 3θ = θ = 45 3θ = : 3θ = : 3θ = : 3θ = 1125 (too high) or 3θ = θ = 315 Given 0 θ 360 => 0 3θ θ = : 3θ = : 3θ = : 3θ = 1395 (too high) => 3θ = 45, 315, 405, 675, 765, : θ = 15, 105, 135, 225, 255, 345 The Dublin School of Grinds Page 75

78 Question 5.8 Step 1: Reference angle: Step 2: cos 1 ( 1 2 ) = 45 Step 3: Step 4: +360: θ 2 = θ 2 = 45 θ 2 = 45 θ = 405 (too high) 2 or Given 0 θ 360 => 0 θ θ = θ 2 = 315 θ = 315 (too high) 2 => θ 2 = 45 2: θ = 90 Question 5.9 Step 1: Reference angle: Step 2: cos 1 ( 1 2 ) = 45 Step 3: θ = θ = 45 or θ = θ = 315 No range => θ = n n Z or => θ = n n Z The Dublin School of Grinds Page 76

79 Question 5.10 a) Step 1: Reference angle: Step 2: 17 cot x = 22 cot x = cos A (but cot A = sin A ) cos x sin x = sin x cos x = tan x = tan 1 ( ) = Step 3: x = x = or x = x = b) 3 sec x 5 = 0 3 sec x = 5 sec x = 5 3 (but sec A = 1 cos A ) => 1 cos x = cos x 1 = 3 Step 1: Reference angle: Step 2: cos x = 3 5 cos 1 ( 3 5 ) = Step 3: x = x = or x = x = The Dublin School of Grinds Page 77

80 Question cot 2 θ = 1 cot 2 θ = 1 2 cot θ = 1 2 (but cot A = cos θ sin θ = 1 2 sin θ cos θ = 2 1 tan θ = 2 cos A sin A ) Step 1: Reference angle: Step 2: tan 1 ( 2) = Step 3: θ = θ = or θ = θ = = π 180 rad = ( π 180 ) rad = π rad 1 = π 180 rad = ( π 180 ) rad = π rad No range => θ = π + 2πn n Z or θ = π + 2πn n Z Question 5.12 M and N are where the graph is equal to zero: => sin 2 θ sin θ = 0 sin θ (sin θ 1) = 0 => sin θ = 0 and sin θ 1 = 0 sin θ = 1 Step 1: Reference angle: sin 1 (0) = 0 Step 2: Step 1: Reference angle: sin 1 (1) = 90 Step 2: Step 3: θ = θ = 0 or θ = θ = 180 Step 3: θ = θ = 90 or θ = θ = 90 No range: => θ = n or θ = n No range: => θ = n ie: θ = 0, 90, 180, 360, 450, 540, etc From diagram: M = 180 and N = 360 (ie: 3rd) (ie: 4th) The Dublin School of Grinds Page 78

81 Question 5.13 P, Q, R, S and T are where the graph is equal zero: => cos θ sin 2θ = 0 => cos θ = 0 and sin 2θ = 0 Step 1: Reference angle: cos 1 (0) = 90 Step 2: Step 1: Reference angle: sin 1 (0) = 0 Step 2: Step 3: θ = θ = 90 or θ = θ = 270 Step 3: 2θ = θ = 0 or 2θ = θ = 180 No range: => θ = n or θ = n No range: => 2θ = n or 2θ = n 2: θ = 180n θ = n ie: θ = 0, 90, 180, 270, 360, 450, 540 etc From diagram: P = 0, Q = 90, R = 180, S = 270, T = 360 (ie: 1st) (ie: 2nd) (ie: 3rd) (ie: 4th) (ie: 5th) Question 6.1 Equipment used: measuring tape and clinometer. (for length) (for angles) w + α β = 180 => w = β x x Using Sine rule: sin α = 15 sin(β α) 15 sin α => x = sin(β α) h sin β = 15 sin α sin(β α) 15 sin α sin β h = sin(β α) β The Dublin School of Grinds Page 79

82 Question 6.2 i) Explanation Measure x with a measuring tape. Measure A and B using a clinometer. Do the following maths: w + B A = 180 w = A B y Using Sine rule: sin B = x sin(a B) x sin B y = sin(a B) x sin B h Using Sine rule: sin A = sin(a B) sin 100 => h = x sin A sin B sin(a B) sin 100 ii) Possible values: x = 20m A = 70 B = sin 70 sin 50 => h = sin(70 50) sin m Question Using Sine Rule: sin y = 5 sin 5 => y = but y is clearly > 90 => Ambiguous case 47m Correct y = = w = 180 => w = x Using Sine Rule: sin = 5 sin 5 => x = => Yes, there is a chance. The Dublin School of Grinds Page 80

83 Question 7.1 Using Cosine Rule: 7 2 = (5)(8) cos θ => θ = 60 Question 7.2 x Using Sine Rule: sin 63 = 20 sin 12 => x = y Using Sine Rule: sin 113 = 20 sin 10 => y = Using Cosine Rule: z 2 = (85.71) 2 + (106.02) 2 2(85.71)(106.02) cos 48 => z 80.16m ie: YZ 80m The Dublin School of Grinds Page 81

84 Question 7.3 i) 25 Using Sine Rule: sin x = 22 sin 60 => x = y = 180 => y = Using Cosine Rule: w 2 = (20) 2 + (18) 2 2(20)(18) cos 48 => w = 13.2 ie: DE = 13.2m ii) As the support moves to the left α will increase. The furthest the support can go is to when it is vertical. After that it will fall over!! sin α = α = sin 1 ( ) 62 α Question 9.1 i) Area of top sector = Area of bottom sector r2 θ = 2 r2 θ ( 1 2 ) (17)2 (0.8) = ( 1 2 ) (10)2 θ ii) => θ = radians Length of arc L = rθ => l 1 = (10)(2.312) = and l 2 = (17)(0.8) 13.6 => l 1 : l 2 = 23.12: 13.6 = 17: 10 The Dublin School of Grinds Page 82

85 Question 9.2 cos x = => x = => θ = Area of a triangle = 1 ab sin C 2 = 1 (3)(3) sin = 4.43 Area of sector = (πr 2 ) ( θ 360 ) = (π)(3) 2 ( ) = 6.27 = = 1.84 m 2 => Volume = (1.84)(4500) = 8280m 3 The Dublin School of Grinds Page 83

86 Question 9.3 a) A = 1 2 r2 θ = ( 1 2 ) (3)2 (2θ) = 9θ m 2 b) tan θ = x 3 => x = 3 tan θ Area of triangle = 1 (base)(perpendicular height) 2 = 1 (3)( 3 tan θ) 2 = 9 tan θ m2 2 c) = 2 ( 9 tan θ) 9θ 2 = 9 tan θ 9θ = 9(tan θ θ)m 2 Question 10.1 Using Pythagoras: y 2 = => y = 10 2 tan x = x = tan 1 ( ) 59 The Dublin School of Grinds Page 84

87 Question 10.2 tan 30 = y 60 y = 20 3 tan x = x = tan 1 ( ) = Using Pythagoras: w 2 = => w = 60 2 Question 10.3 a) Using Pythagoras: y 2 = => y = 74 Using Pythagoras: x 2 = ( 74) 2 => x = 9.11 cm b) tan w = 3 74 w = tan 1 ( 3 74 ) The Dublin School of Grinds Page 85

88 Question 10.4 Using Pythagoras: x 2 = => x = cm tan 6 = 2.5 y => y = Using Pythagoras: w 2 = => w = Question 11.1 The Dublin School of Grinds Page 86

89 Question 11.2 a) X 0 2X 0 π 4 π 2 sin(2x) sin(2x) π 2 π 3π 4 3π 2 π 2π b) c) Range= [ 3, 3] Period = π radians Question 11.3 x sin x sin x The Dublin School of Grinds Page 87

90 Question 11.4 i) Sine: because at the y axis, sine graphs will be at the midway line. ii) The graph must be in the form a + b sin cx a is the vertical translation: => a = 3 b is the amplitude: => b = 8 (This is positive since the graph is not inverted) Period = 360 c From the graph, the distance along the x axis from A to B is 3 of a period. Therefore period = => 360 c = 120 => c = 3 So the equation of the graph is y = sin 3x. iii) The distance along the x axis from B to C is 1 of a period => C: (90, y) but y = sin 3(90) 2 => C: (90, 5) The distance along the x axis from C to D is 3 of a period => D: (180, y) but y = sin 3(180) 4 => D: (180, 3) Question 11.5 i) Woodmice: 14,500 Owls: 9,000 ii) Woodmice must be a cosine function, because at the y axis, cosine graphs will be at a maximum or a minimum. The graph must be in the form a ± b cos x a is the vertical translation: => a = 11 b is the amplitude: => b = 3.5 (this is positive because the graph is not inverted) period = 360 c From the graph, the period is 12 => 360 c = 12 => c = 30 So the equation of the graph is y = cos 30x iii) Owls must be a sine function, because at the y axis, sine graphs will be at the midway line. The graph must take the form a ± b sin cx a is the vertical translation: => a = 9 b is the amplitude: => b = 2 (this is positive because the graph is not inverted) period = 360 c From the graph, the period is 12 The Dublin School of Grinds Page 88

91 => 360 c = 12 => c = 30 So the equation of the graph is y = sin 30x iv) Owls was higher than woodmice from: 3 7 in Year 1 => for 4 months v) Greatest difference seems to have been at 11 months: Owls 8,000 Woodmice 14,000 => difference 6,000 Question 11.6 At P: y = 1 2 ie: sin 2x = 1 2 Step 1: Reference angle: Step 2: sin 1 ( 1 2 ) = 30 Step 3: Step 4: 2x = x = 30 2x = : 2x = : 2x = 750 (too high) Given 0 x => 0 2x 720 or 2x = x = 150 2x = : 2x = : 2x = 870 (too high) => 2x = 30, 150, 390, 510 2: x = 15, 75, 195, 255 Looking at the graph P is roughly 3 8 between π and 2π, ie: 3 between 180 and 360, which => P = 255 Change to radians: 1 = π = 255π 180 = 17π 12 P = ( 17π 12, 1 2 ) The Dublin School of Grinds Page 89

92 Question 13.1 A + B A B sin A + sin B = 2 sin cos => sin 10 + sin 80 = 2 sin cos 2 2 = 2 sin 45 cos( 35) but cos( A) = cos A => sin 10 + sin 80 = 2 sin 45 cos 35 = 2 ( 1 ) cos 35 2 = 2 cos 35 Question 13.2 A + B sin A + sin B = 2 sin 2 2 3A + A => sin 3A + sin A = 2 sin cos 2 = 2 sin 2A cos A A B cos A + B cos A + cos B = 2 cos 2 2 3A + A => cos 3A + cos A = 2 cos cos 2 = 2 cos 2A cos A 3A A 2 A B cos 3A A 2 => sin 3A + sin A cos 3A + cos A becomes 2 sin 2A cos A 2 cos 2A cos A = tan 2A Question cos A cos B = cos(a + B) + cos(a B) cos A cos B = 1 2 cos(a + B) + 1 cos(a B) 2 cos 3x cos 2x = 1 2 cos(3x + 2x) + 1 cos(3x 2x) 2 = cos 5x 2 + cos x 2 The Dublin School of Grinds Page 90

93 Solutions to past and probable exam questions Question 1 i) Using Pythagoras: x 2 = x 2 = 5 x = 2.236m ii) y tan y = 2 1 y = tan 1 (2) = 63.4 w = 180 w = 53.2 ie: EGF 53 p Using Sine Rule: sin 63.4 = 0.5 sin 53.2 => p = ie: GE = but DE = m (from (i)) => DG = m Comment: Nothing major here. Question 2 i) cos 35 = 3 x => x = 3.66 m ii) Using Cosine Rule: y 2 = (2) 2 + (1.83) 2 2(2)(1.83) cos 35 => y = The Dublin School of Grinds Page 91

94 Using Cosine Rule: w 2 = (2) 2 + (3.66) 2 2(2)(3.66) cos 35 => w = => Total Length = 2(3.66) + 2(1.163) + 2(2.324) m Comment: Fine question. Roof trusses seem to be in fashion since Project Maths was introduced so watch out for one of these appearing again. Question 3 Using Cosine Rule: 4 2 = (5)(4) cos x => x = Using Cosine Rule: 7 2 = y 2 2(5)(y) cos y y 24 = 0 a = 1 b = 6.25 c = 24 Comment: Handy. y = b ± b2 4ac 2a y = ( 6.25) ± ( 6.25)2 4(1)( 24) 2(1) y = 8.94 or y = 2.69 (ignore) ie: BC = 8.94 but BD = 4 => DC = The Dublin School of Grinds Page 92

95 Question 4 i) Using Cosine Rule: c 2 = p 2 + d 2 2pd cos α => cos α = c2 p 2 d 2 2pd Using Cosine Rule: b 2 = q 2 + d 2 2qd cos β => cos β = b2 q 2 d 2 2qd ii) We know cos α = c2 p 2 d 2 and cos β = b2 q 2 d 2 2pd 2qd but β = 180 α => cos(180 α) = b2 q 2 d 2 2qd (but cos(a B) = cos A cos B + sin A sin B) cos 180 cos α + sin 180 sin α = b2 q 2 d 2 cos α = b2 q 2 d 2 2qd 1: cos α = b2 q 2 d 2 2qd 2qd Putting cos α = cos α c 2 p 2 d 2 = b2 q 2 d 2 2pd 2qd 2qd(c 2 p 2 d 2 ) = 2pd(b 2 q 2 d 2 ) 2qdc 2 2qdp 2 2qd 3 = 2pdb 2 + 2pdq 2 + 2pd 3 2d: qc 2 qp 2 qd 2 = pb 2 + pq 2 + pd 2 pb 2 + qc 2 = qp 2 + pd 2 + pq 2 + qd 2 pb 2 + qc 2 = p(pq + d 2 ) + q(pq + d 2 ) b 2 + qc 2 = (p + q)(pq + d 2 ) Comment: Tough question. Quite old school. Would separate the top students from the rest. The Dublin School of Grinds Page 93

96 Question 5 a) Using Pythagoras: w 2 = => w = 25 tan x = 7 24 x = Using Cosine Rule: 12 2 = (20)(25) cos y => y = but α = x + y = = 44 Using Cosine Rule: 25 2 = (20)(12) cos β => β = 100 b) Length of arc: L = 2πr ( θ 360 ) Increase of 1 in α: L = (2)(π)(25) ( ) = 0.138π Increase of 1 in β: L = (2)(π)(12) ( ) = 0.06π => Answer = α c) It can be seen from the formulae used in (b) that the error depends on the radius: => When PR > QR there will be a greater error in α. and... When QR > PR there will be greater error in β. The Dublin School of Grinds Page 94

97 d) Comment: Scandalous question. Stupid, stupid, stupid! Question 6 (a) (i) (ii) Using Sine Rule: x sin 16.3 = 100 sin 3.7 => x = sin 20 = h => h 149m b) Using Pythagoras: x 2 = => x 2 = => x = => FC = The Dublin School of Grinds Page 95

98 c) Using Pythagoras: y 2 = ( ) 2 => y 221m d) Using Pythagoras: = w => w = e) => total length = m Area of each triangle = 1 2 (base)(perpendicular) b 2 = ( 1 2 a) 2 + x 2 b 2 = a2 4 + x2 b 2 a2 4 = x The Dublin School of Grinds Page 96

99 = 1 2 (a) ( b 2 a2 4 ) = a 2 b 2 a2 4 Surface area = 4 triangles = 4 ( a 2 b 2 a2 4 ) f) (i) (ii) = 2a b 2 a2 4 = 2(230) (221) 2 (230)2 4 = 86,812m 2 Area of casing stone = (0.86)(0.86) = m 2 => Number of casing stones = 86, = 117,000 Comment: Nice question with lots going on. Question 7 a) sin 30 = x 10 => x = 5m b) From Diagram above: Using Pythagoras: 10 2 = y 2 c) => y = 5 3 => New distance from wall = m tan α = α = tan 1 ( ) α = d) From Diagram above: Using Pythagoras: w 2 = => w = 7.55m => protruding part = = 2.45m Comment: Nothing major here. Nice. The Dublin School of Grinds Page 97

100 Question 8 a) (i) r 1 = 16cm r 2 = 9cm (ii) = = 25cm b) (i) Using Pythagoras: 25 2 = x 2 => x = 24cm (ii) From above diagram: cos w = 7 25 => w = cos 1 ( 7 25 ) = radians Area of sector = 1 2 r2 θ = 1 2 (16)2 (1.287) = 164.7cm 2 c) (i) ie: (ii) Using Pythagoras: (16 + r) 2 = f 2 + (16 r) r + r 2 = f r + r 2 64r = f 2 ie: => f = 8 r Using Pythagoras: (9 + r) 2 = p 2 + (9 r) r + r 2 = p r + r 2 36r = p 2 => p = 6 r The Dublin School of Grinds Page 98

101 d) f + p must equal 24 ie: 8 r + 6 r = r = 24 r = 12 7 r = cm e) a + b Area of trapezium = ( ) h [from pg 8 of log tables] 2 = ( ) (24) 2 = 300cm 2 f) Area between wheels = Trapezium Sector APC Sector CQB Small wheel = r2 θ πr 2 = (9)2 (1.855) π ( ) π radians in a triangle: π + α = π 2 => α = radians => CQB = π 2 = rad => Area between wheels = 33cm 2 Now area of wheels = πr 2 + πr 2 + πr 2 = π(16) 2 + π(9) 2 + π ( ) 2 = cm 2 => % = % Comment: Wow! This would cause problems. Project Maths at its finest. The Dublin School of Grinds Page 99

102 Question 9 a) Max value of cos is 1. Min value of cos is 1. b) (i) t(s) 0 π 12 π 6 => SP = (1) = 120 mm Hg DP = ( 1) = 80 mm Hg π 4 P(mm Hg) (ii) π 3 5π 12 π 2 7π 12 2π 3 (iii) Range = [80, 120] Period = π 3 (iv) Duration is the period, ie: π seconds 3 Number of beats per minute 60 π 3 = 57 c) (i) Marked on the graph with a point. => π 18, 5π 18, 7π 18, 11π 18 (ii) 110 = cos 6t 10 = 20 cos 6t 1 = cos 6t 2 cos 6t = 1 2 Step 1: Reference angle: Step 2: cos 1 ( 1 2 ) = 60 Step 3: Step 4: 6t = t = 60 or 6t = t = 300 6t = 60/ 420 / 780 / 1140 /etc. or 6t = 300 / 660 / 1020 / 1380 / etc. => 6t = 60 / 300 / 420 / 660 / 780 / 1020 / 1140 / 1380/ etc 6: t = 10 / 50 / 70 / 110 / 130 / 170 / 190 / 230 / etc 1 = π 180 radians => t = π 18 / 5π 18 / 7π 18 / 11π 18 / 13π 18 / 17π 18 / 19π 18 / 23π 18 / etc Instead of using this method you could have subbed your values from part (i) into the formula and shown that the left equals the right. The Dublin School of Grinds Page 100

103 d) This part of the question requires differentiation: P = cos 6t dp = 20( sin 6t) (6) dt = 120 sin 6t Beginning (t = 0): One fifth (t = π ): 15 Three fifths (t = π ): 5 End (t = π ): 3 dp = 120 sin 6(0) dt = 0 mm Hg/sec dp dt = 120 sin 6 ( π 15 ) = mm Hg/sec dp dt = 120 sin 6 (π 5 ) = 70.5 mm Hg/sec dp dt = 120 sin 6 (π 3 ) = 0 mm Hg/sec Comment: Awkward wording but this is the sort of stuff the Examiner likes these days. The Dublin School of Grinds Page 101

104 Question 10 (a) (i) Rhombus (ii) Square (Note: the lengths would have to change) (b) (i) Perimeter: 2a + 2b (ii) 1. Diagonals Kite = ABC + ACD = 1 2 (p) (1 2 q) (p) (1 2 q) = 1 2 pq (c) (i) Using Cosine Rule: AC 2 = (30) 2 + (40) 2 2(30)(40) cos 100 => AC = 54 cm 2. Sides Kite = ABC + ACD = 1 2 ab sin θ + 1 ab sin θ 2 = ab sin θ (ii) Usiing Sine Rule: 40 sin x = 54 sin 100 => x 46.8 (iii) From diagram above: y = 33.2 sin 33.2 = x 40 => x = 21.9 => BD = 2(21.9) = 43.8 m (iv) Kite = ABC + ADC = 1 2 (30)(40)(sin 100) (30)(40)(sin 100) = 1182 cm 2 The Dublin School of Grinds Page 102

105 (d) (e) (i) Using Pythagoras: d 2 = ( d 2 ) 2 + (f) 2 d 2 = d2 4 + f2 3d 2 4 = f2 d 3 2 = f (ii) => d f becomes d d 3 2 Now tan w = d d 3 2 d 2 d d tan w = ( ) 1 2 w = tan 1 (2 3) w = 15 d d sin g = d 2 d g = sin 1 ( 1 2 ) g = 30 => ADC = 75 BAD = 150 ABC = 75 BCD = 60 The Dublin School of Grinds Page 103

106 (f) tan 15 = tan(45 30) tan A tan B but tan(a B) = 1 + tan A tan B tan 45 tan 30 => tan(45 30) = 1 + tan 45 tan = (1) ( 1 3 ) = (1)( 3) (1)(1) 3 (1)( 3) + (1)(1) 3 = = = = = 2 3 Comment: When broken down it s actually okay, but it s easy to see how students would struggle here. Question 11 (a) tan 60 = h x => x = h tan 60 = h 3 (b) tan 30 = h y => y = h tan 30 = h 3 Comment: Standard question. Nice. Using Pythagoras: (h 3) 2 = (24) 2 + ( h 3 ) 2 3h 2 = h2 3 8h 2 3 = 576 => h 14.7 m The Dublin School of Grinds Page 104

107 Question 12 (a) (b) (c) a 2 = b 2 + c 2 2bc Cos A b 2 = a 2 + c 2 2ac Cos B c 2 = a 2 + b 2 2ab Cos C BD = DC Using Cosine Rule: 7 2 = (2x) 2 2(4)(2x) Cos y 49 = x 2 16x Cos y 4x 2 16x Cos y = 33 1 Using Cosine Rule again: = x 2 2(4)(x) Cos y = 16 + x 2 8x Cos y x 2 8x Cos y = Comment: Nice short question. 1 4x 2 16x Cos y = x x Cos y = 7.5 2x 2 = 40.5 x 2 = x = 4.5 => BC = 9 Question 13 (a) cos 3x has a range of [ 1, 1] and period cos 3x has a range of [ 2, 2] and period cos 2x has a range of [-3, 3] and period 180. (b) See above Comment: Easy peeazy lemon squeezy. The Dublin School of Grinds Page 105

108 Question 14 (a) x 0 π π 3π π 5π 3π 7π 2π cos x cos 3x (b) At P: cos 3x = 1 Step 1: Reference angle: Step 2: cos 1 (1) = 0 Step 3: No range: 3x = x = 0 or 3x = x = 360 3x = n or 3x = n => x = n x = n => x = 0, 120, 240, 360, 480, 600 Now P is roughly a third of the way between π and 2π. i.e. roughly a third between 180 and Ha, this is exactly one of the answers we got! 1 = π 180 rad 240 = 240 ( π 180 ) rad Comment: Fine question. These are getting popular. = 4π 3 rad i. e: P = ( 4π 3, 1) The Dublin School of Grinds Page 106

109 Question 15 Graph 1: Function: 2 sin x Period: 360 Range: [ 2, 2] Graph 2: Function: 3 cos 2x Period: 180 Range: [ 3, 3] Graph 3: Function: 4 cos 3x Period: 180 Range: [ 4, 4] Graph 4: Function: tan x Period: 180 Range: No range Graph 5: Function: 2 sin x 2 Period: 720 Range: [ 2, 2] The Dublin School of Grinds Page 107

110 Graph 6: Function: 2 + sin 3x Period: 120 Range: [1, 3] Graph 7: Function: 1 sin 5x 2 Period: 72 Range: [ 1 2, 1 2 ] Graph 8: Function: cos 4x Period: 90 Range: [ 2, 4] Comment: Again, this type of question is getting popular. Know it! The Dublin School of Grinds Page 108

111 Question 16 At W, X and Y the graph = 0 2 sin 3θ cos 2θ + sin 3θ = 0 sin 3θ (2 cos 2θ + 1) = 0 sin 3θ = 0 Step 1: Reference angle: sin 1 (0) = 0 Step 2: 2 cos 2θ + 1 = 0 cos 2θ = 1 2 Step 1: Reference angle: cos 1 ( 1 2 ) = 60 Step 2: Step 3: 3θ = θ = 0 No range: 3θ = n => θ = 120n or 3θ = θ = 180 or 3θ = n θ = n Step 3: 2θ = θ = 120 No range: 2θ = n => θ = n or 2θ = θ = 240 or 2θ = n θ = n From diagram: => 120 = 120 ( π 180 ) rad = 2π 3 rad => θ = 0, 60, 120, 180, 240, 300, 360, 420, etc W X Y (ie: 3 rd ) (ie: 6 th )(ie: 7 th ) 1 = π 180 rad 300 = 300 ( π 180 ) rad = 5π 3 rad 360 = 360 ( π 180 ) rad = 2π rad Comment: Fine question once you know how to start. Question 17 (a) (i) (ii) Period = 12 hours Range = [3000, 13,000] It s a cosine graph because at the y-axis (when x = 0), cosine graphs will be at a maximum or minimum. => The graph must be in the form a + b cos cx a is the vertical translation: => a = 8 b is the amplitude: => b = 5 (This is negative because the graph is inverted, ie: cos should start downhill, but this starts uphill) Period = 360 c => 12 = 360 c => c = 30 So the equation of the graph is y = 8 5 cos 30x (iii) 9,300 (iv) 22: 30 and 01: 30 The Dublin School of Grinds Page 109

112 (v) (vi) 6am No, because in the middle of the night most people will be in bed. (b) (i) The max value for sin is 1. This happens at sin 90, sin 450, sin 810 etc. ie: every rotation, ie: every 2π Looking at P(t) = 30 sin 2πt + 120, we see this would be every second. => 60 heartbeats per minute. (ii) t P (iii) => Arthur is in the normal range. Comment: This question could get blood pressure going. Nice question. The Dublin School of Grinds Page 110

113 Question 18 (a) (i) (ii) (iii) (iv) (v) ie: h = sin 45t t h The height is periodic. Max = 415cm Min = 385cm Period = 8 seconds => 60 = 7.5 times per minute 8 => Yes He would need the period to be 10 seconds. period = 360 c 10 = 360 c => c = 36 => The function of the movement of the boat is 15 sin 36t (vi) period = = 6 ie: the period is now less. ie: it is more wavey. Reason: Perhaps the weather or location. (b) (i) (ii) (iii) q is the midpoint line: p is the amplitude: 60m 2π π 16 = 32 => q = 75 => p = 60 (iv) So our formula is height = 60 cos 12x + 75 Now let s find the times that a given capsule is at the top. So we have: 135 = 60 cos 12x = 60 cos 12x cos 12x = 1 The Dublin School of Grinds Page 111

114 Step 1: Reference angle: Step 2: cos 1 (1) = 0 Step 3: 12x = x = 0 or 12x = x = 360 (v) => 12x = 0, 360, 720, etc. x = 0, 30, 60, etc. ie: a full rotation every 30 minutes. π 3 Comment: A real Project Maths question. Tough. The Dublin School of Grinds Page 112

115 Question 19 (a) (b) (i) (ii) a 2 = b 2 + c 2 2bc cos A (150) 2 = (120) 2 + (134) 2 2(120)(134) cos x = cos x 9856 = cos x cos 1 ( ) = x = x Area = 1 ab sin C 2 = 1 (120)(134) sin = 7653 m 2 Comment: Fine. Question 20 (a) (b) Arc length AB: l = rθ l = (x + 0.6)(θ) = xθ + 0.6θ cos(a + B) = cos A cos B sin A sin B cos(a + A) = cos A cos A sin A sin A cos 2A = cos 2 A sin 2 A xθ + 0.6θ + 3 = xθ + 1.8θ 3 = 1.2θ = θ 2.5 radians = θ Arc length CD: l = rθ l = (x + 1.8)(θ) = xθ + 1.8θ Comment: Nothing too hard here at all. The Dublin School of Grinds Page 113

116 Question 21 (a) (i) [ 311, 311] (ii) Period every 0.02 seconds. => 50 period per second. (b) (i) (c) (ii) = (on calculator) 220 (i) kσ = V max k = V max σ = = (ii) In the form V = a sin(bt): Period = 2π b => b = 2π period If it has 60 complete periods every second, then the period is 1 => b = 2π ( 1 60 ) = 120π In the form V = a sin(bt) Range = [ a, a] but a = V max and from above V max = kσ = (1.414)(110) = Comment: A bit weird at times, especially the last few parts. Okay apart from that. 60 The Dublin School of Grinds Page 114

117 Question 22 (a) sin(a + B) tan(a + B) = cos(a + B) sin A cos B + cos A sin B = cos A cos B sin A sin B Now just divide everything by cos A cos B sin A cos B cos A sin B + = cos A cos B cos A cos B cos A cos B sin A sin B cos A cos B cos A cos B tan A + tan B = 1 tan A tan B (b) Step 1: Reference angle: sin 1 ( 3 2 ) = 60 Step 2: Step 3: 3x = x = 60 or 3x = x = 120 Step 4: Given: 0 x 360 3: 0 3x 1080 => 3x = 60/420/780 or 3x = 120/480/840 => 3x = 60/120/420/480/780/840 3: x = 20/40/140/160/260/280 Comment: Free marks! Question 23 (a) Using Pythagoras: (20 73) 2 = (8r) 2 + (3r) = 64r 2 + 9r = 73r = r 2 r = 20cm (b) Area of rectangle: Area of triangle: (8r)(r) i. e.: (160)(20) = 3200 = 1 2 bh = 1 2 (8r)(3r) i. e. : 1 2 (120)(60) = 4800 => Area of quadrilateral = 8000cm 2 The Dublin School of Grinds Page 115

118 (c) (i) tan x = x = tan 1 ( ) = => HAP = (ii) The two quadrilaterals equal 16000cm 2. Next is the big sector: Next is the small sector: A = πr 2 ( θ 360 ) A = π(80) 2 ( ) A = A = πr 2 ( θ 360 ) A = π(20) 2 ( ) A = => Total area = 28833cm 2 Comment: Is this copied from one of our teachers at The Dublin School of Grinds?! Question = ABT = => KBQ = (a) sin x = x = sin 1 ( ) x = => α = 11.5 (b) (c) (i) Sub t = 0: (ii) At B, h = 0: a 2 = b 2 + c 2 2bc cos A x 2 = (190) 2 + (385) 2 2(190)(385) cos 18 x 2 = 45, x 213m h = 6(0) (0) + 8 h = 8m 0 = 6t t + 8 6t 2 22t + 8 = 0 2: 3t 2 11t 4 = 0 (3t + 1)(t 4) = 0 t 1 t = 4 3 => OB = (4)(38) = 152 tan x = x = tan 1 ( ) x = 3 The Dublin School of Grinds Page 116

119 (d) (i) Also, (ii) Using Pythagoras: tan θ = h d d = h tan θ d = h ( 1 2 ) d = 2h CD + h = 25 => CD = 25 h (25) 2 = (d) 2 + ( CD ) 2 (25) 2 = (2h) 2 + (25 h) = 4h h + h 2 5h 2 50h = 0 5h: h 10 = 0 h = 10m Comment: Quite basic. Applied Maths students were at an advantage for part (c). The Dublin School of Grinds Page 117

120 Co-ordinate Geometry of the Line Co-ordinate geometry of the line is worth 4% to 9% of the Leaving Cert. It appears on Paper 2. 1) Distance between two points PQ = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 NOTE: This formula appears on Page 18 of The Formulae & Tables Booklet 2) Midpoint of a line Midpoint = ( x 1+x 2 2, y 1+y 2 ) 2 NOTE: This formula appears on Page 18 of The Formulae & Tables Booklet 3) Slope of a line A line going up from left to right has a positive slope: + A line going down from left to right has a negative slope: NOTE: Horizontal lines have a slope of 0 NOTE: Vertical lines have a slope of infinity There are 6 ways to find the slope of a line: Way (i) if given 2 points we use the formula m = y 2 y 1 x 2 x 1 NOTE: This formula appears on Page 18 of The Formulae & Tables Booklet The Dublin School of Grinds Page 118

121 Way (ii) if given its equation in the form ax + by + c = 0 slope = a b Way (iii) if given its equation in the form y = mx + c slope = m Way (iv) if given its equation in the form y y 1 = m(x x 1 ) slope = m Way (v) angle of inclination: If we are given a line which makes an angle α, measured anti-clockwise from the positive side of the x-axis, then: m = Tan α Way (vi) the Rise Run method. FORMULA: Slope = Rise Run The Rise is the change in the y value (this can be positive or negative), as we read from left to right. The Run is the change in the x value (this is always positive). Example 1 Draw the line through the point (2, 1), with a slope 3 5. Solution: Rise Run = 3 5 y value increases from 1 to 4 x value increases from 2 to 7 New point = (7, 4) The Dublin School of Grinds Page 119

122 4) Slopes of parallel lines Parallel lines have equal slopes: m 1 = m 2 5) Slopes of perpendicular lines To find the slope of a perpendicular line you turn the slope upside down and change the sign: eg eg To show two lines are perpendicular we multiply their slopes and show they are equal to 1 (m 1 m 2 = 1) 6) The equation of a line There are 3 ways we can find the equation of a line: 1) Use the formula y y 1 = m(x x 1 ) NOTE: This formula appears on Page 18 of The Formulae & Tables Booklet 2) The general form of the equation is: ax + by + c = 0 A line parallel to the general form would be : ax + by + d = 0 A line perpendicular to the general form would be: bx ay + d = 0 i.e. Swap the letters in front of x and y and then change the sign of whatever is in front of the y. 3) Equation in the form y = mx + c m stands for the slope. c stands for the y-intercept (where the line cuts the y-axis). 7) The point of intersection of a line and the axes When a line cuts the x-axis: y = 0 When a line cuts the y-axis: x = 0 8) Point of intersection of two lines RULE: Use simultaneous equations NOTE: These are covered in the Algebra chapter The Dublin School of Grinds Page 120

123 9) The area of a triangle We know from basic area and volume questions that: area of triangle = 1 (base)(perpendicular height) 2 In trigonometry we often did not know the perpendicular height so we used another formula: area of triangle = 1 2 absinc NOTE: This formula appears on page 16 of the formulae and tables booklet. However in co-ordinate geometry we often do not have the information required for the two formulae above. Therefore we have another formula: area of triangle = 1 2 x 1y 2 x 2 y 1 NOTE: This formula appears on Page 18 of The Formulae & Tables Booklet. To use this formula we must move one vertex (which is a fancy name for the corner of the triangle) to (0, 0) and translate the other two points. 10) Perpendicular distance from a point to a line The shortest distance from any point to a line is the perpendicular distance. To find this we use the formula: d = ax 1 + by 1 + c a 2 + b 2 NOTE: This formula appears on Page 19 of The Formulae & Tables Booklet The x 1 and y 1 come from the point (x 1, y 1 ), while the a, b and c come from the line ax + by + c = 0 Be careful when dealing with the modulus brackets in these questions. This is covered in the Algebra chapter. 11) The angle between two lines To find the acute angle (θ) between two lines with slopes m 1 and m 2 we use the formula: Tanθ = ± m 1 m m 1 m 2 NOTE: This formula appears on Page 19 of The Formulae & Tables Booklet To find the obtuse angle between the lines we calculate 180 θ 12) Dividing a line segment in a ratio If P (x 1, y 1 ) and Q (x 2, y 2 ) are two points we can divide [PQ] in the ratio a: b, using the point R. Internal division: NOTE: This formula appears on Page 18 of The Formulae & Tables Booklet The Dublin School of Grinds Page 121

124 13) Past and probable exam questions Question 1 The Dublin School of Grinds Page 122

125 Question 2 Question 3 The Dublin School of Grinds Page 123

126 Question 4 Question 5 (a) Find the area of the parallelogram whose vertices are (-1,3), (0,2), (5,4) and (4,5). (b) The line B contains the points (6,-2) and (-4,10). The line A with equation ax + 6y + 21 = 0 is perpendicular to B. Find the value of the real number a. (c) N is the line tx + (t 2)y + 4 = 0, where t R. (i) Write down the slope of N in terms of t. (ii) Given that the angle between N and the line x 3y + 1 = 0 is 45, find the two possible values of t. Question 6 (a) A(2t,0) and B(0,-t) are two points. If AB = 20, find the two values of t. (b) Find the equations of the lines through the point (4,3) which make an angle of 45 with the line 6x + y 5 = 0. (c) Find the co-ordinates of the circumcentre of the triangle whose vertices are (4,6), (-4,-2) and (10,0). The Dublin School of Grinds Page 124

127 Question 7 (a) Write down the slope of the line k in the given diagram. Hence, find the equation of k in the form ax + by + c = 0. (b) Find the equations of the two lines through the point (1,-1) which make an angle of 45 with k. Question 8 (a) (b) Question 9 The Dublin School of Grinds Page 125

128 Question 10 Question 11 The Dublin School of Grinds Page 126

129 Question 12 Question 13 Question 14 The Dublin School of Grinds Page 127

130 Question 15 The Dublin School of Grinds Page 128

131 14) Solutions to Co-ordinate Geometry of the Line Question 1 (a) Equation Line x + 2y = 4 l 2x y = 4 m x + 2y = 8 j 2x y = 2 n Workings: Write lines in the form y = mx + c 1 st line: 2y = x 4 y = 1 2 x 2 => slope = 1 2 / y intercept = 2 2 nd line: y = 2x 4 y = 2x + 4 => slope = 2 / y intercept = 4 3 rd line: 2y = x + 8 y = 1 2 x + 4 => slope = 1 2 / y intercept = 4 b) 4 th line: y = 2x + 2 y = 2x 2 => slope = 2 / y intercept = 2 Workings: Use workings from part (a) and also find where any line crosses the x-axis, let s use j: x + 2(0) = 8 x = 8 (c) Remaining line is k, Points (4, 0) (0, 2) x 1 y 1 x 2 y 2 y y 1 = m(x x 1 ) y 0 = 1 (x 4) 2 y = 1 2 x x + y 2 = 0 2 Comment: No worries. m = y 2 y 1 x 2 x 1 m = m = 1 2 The Dublin School of Grinds Page 129

132 Question 2 (a) Sub in: 3(4k 2) 4(3k + 1) + 10 = 0 12k 6 12k = 0 0 = 0 => on line (b) slope of l 1 = a b = 3 4 = 3 4 => slope of l 2 = 4 3 Equation of l 2 : y y 1 = m(x x 1 ) y (3k + 1) = 4 (x (4k 2)) 3 y 3k 1 = 4 (x (4k 2)) 3 3: 3y 9k 3 = 4x + 16k 8 4x + 3y 25k + 5 = 0 (c) Sub in: 4(3) + 3(11) 25k + 5 = k = 0 50 = 25k => k = 2 (d) l 2 goes through Q when k = 2 => equation of l 2 : 4x + 3y 25(2) + 5 = 0 4x + 3y 45 = 0 4 l 1 : 12x 16y + 40 = 0 3 l 2 : 12x 9y = 0 25y = 0 => y = 7 => x = 6 => (6, 7) Comment: A bit weird but nothing you can t handle. The Dublin School of Grinds Page 130

133 Question 3 (a) (b) y y 1 = m(x x 1 ) Point (0, 10) x 1 y 1 Slope = tan α = tan(90 θ) = tan (90 tan 1 ( 4 3 )) = 3 4 [on calculator] => y ( 10) = 3 (x 0) 4 y + 10 = 3 4 x 4: 4y + 40 = 3x 3x 4y 40 = 0 (c) Shortest distance is always perpendicular distance: d = ax 1 + by 1 + c a 2 + b 2 x 1 = 0 y 1 = 0 a = 3 b = 4 c = 40 (3)(0) + ( 4)(0) + ( 40) d = (3) 2 + ( 4) 2 d = d = 8km (d) Time = Distance Speed Using the left triangle from our diagram: 10 2 = y => Time = 6 15 hours = 24 minutes => Time is 12: 24pm => y = 6km (e) Re-drawing a sketch of the situation: 9 2 = w 2 => w = 17 => ship visible for 2 17 km Time = Distance Speed = hours = minutes Comment: A nice modern question, which mixes a few topics. Not the easiest however. The Dublin School of Grinds Page 131

134 Question 4 (a) y = mx + c y = 3x + 2 (b) See diagram to the right: (c) See diagram to the right: (d) Slope of l 1 = 3 => Slope of l 4 = 1 3 Equation of l 4 : y y 1 = m(x x 1 ) y 4 = 1 (x 0) 3 y 4 = 1 3 x 3: x + 3y 12 = 0 Sub (27, 4): ( 4) 12 = => No Comment: Very handy. Is this real life? Question 5 a) Parallelogram is 2 triangles. Area of triangle = 1 2 x 1y 2 x 2 y 1 (0, 2) ( 1, 3) (4, 5) (0,0) ( 1,1) (4,3) => Area of triangle = 1 ( 1)(3) (4)(1) 2 = 7 2 => Area of parrallelogram = (2) ( 7 2 ) = 7 units 2 b) Slope of A = a b = a 6 Slope of B: m = y 2 y 1 x 2 x 1 x 1 y 1 x 2 y 2 (6, 2) ( 4, 10) 10 ( 2) m = 4 6 m = 6 5 The Dublin School of Grinds Page 132

135 c) (i) If perpendicular: ( a 6 ) ( 6 5 ) = 1 m 1 m 2 = 1 6a 30 = 1 6a = 30 a = 5 (ii) Slope of N: t t 2 Slope = a b = t t 2 Slope of new line: = a b = 1 3 = 1 3 Angle between two lines: tan θ = ± m 1 m m 1 m 2 tan 45 = ± ( t t 2 ) (1 3 ) 1 = ± 1 + ( t t 2 ) (1 3 ) ( t)(3) (1)(t 2) (3)(t 2) (1)(3)(t 2) + ( t)(1) (3)(t 2) 3t t = ± 3t 6 t 1 = ± 4t + 2 2t 6 2t 6 = ±( 4t + 2) 2t 6 = 4t + 2 6t = 8 t = 4 3 or 2t 6 = 4t 2 4 = 2t 2 = t Comment: Messy but fine. The Dublin School of Grinds Page 133

136 Question 6 a) b) AB = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 20 = (0 2t) 2 + ( t 0) 2 => 20 = 4t 2 + t 2 20 = 5t 2 4 = t 2 => t = ± 2 Slope of 6x + y 5 = 0 is a b = 6 1 = 6 Call the slope of the other line w. Angle between two lines: tan θ = ± m 1 m m 1 m 2 ( 6) (w) tan 45 = ± 1 + ( 6)(w) x 1 y 1 x 2 (2t, 0 ) ( 0, t y 2 ) 6 w 1 = ± 1 6w 1 = ± 4t + 2 2t 6 1 6w = ±( 6 w) 1 6w = 6 w 7 = 5w 7 5 = w Equation of line : y y 1 = m(x x 1 ) y 3 = 7 (x 4) 5 y 3 = 7 28 x 5 5 5: 5y 15 = 7x 28 7x 5y 13 = 0 or or 1 6w = 6 + w 5 = 7w 5 7 = w y 3 = 5 (x 4) 7 y 3 = 5 20 x : 7y 21 = 5x x + 7y 41 = 0 c) Circumcentre is where the perpendicular bisectors meet. Sketch: The Dublin School of Grinds Page 134

137 x 1 y 1 x 2 y 2 Midpoint of ( 4, 2 ) to ( 4, 6 ): x 1 y 1 x 2 y 2 Slope of ( 4, 2 ) to ( 4, 6 ): = ( x 1 + x 2 2 = ( = (0, 2), y 1 + y 2 ) 2, ) 2 = y 2 y 1 x 2 x 1 = 6 ( 2) 4 ( 4) = 1 => Slope of perpendicular bisector = 1 => Equation of perpendicular bisector: y y 1 = m(x x 1 ) y 2 = 1(x 0) y 2 = x x + y 2 = 0 1 x 1 y 1 x 2 y 2 Midpoint of (4, 6 ) to (10, 0 ): x 1 y 1 x 2 y 2 Slope of (4, 6 ) to (10, 0 ) => Slope of perpendicular bisector = 1 => Equation of perpendicular bisector: Comment: Not nice, but nothing you don t know. = ( x 1 + x 2 2 = ( = (7, 3) = y 2 y 1 x 2 x 1 = = 1, y 1 + y 2 ) 2, ) y y 1 = m(x x 1 ) y 3 = 1(x 7) y 3 = x 7 x y 4 = x + y 2 = 0 2 x y 4 = 0 2x 6 = 0 x = 3 => y = 1 => Circumcentre: (3, 1) The Dublin School of Grinds Page 135

138 Question 7 a) Slope = Rise Run = 3 4 (0, 2) (4, 5) y y 1 = m(x x 1 ) y 2 = 3 (x 0) 4 y 2 = 3 (x 0) 4 b) y 2 = 3 4 x 4: 4y 8 = 3x 3x 4y + 8 = 0 Slope of k = a b = 3 4 Call slope of other line w = 3 4 Angle between two lines: tan θ = ± m 1 m m 1 m 2 tan 45 = ± (3 4 ) (w) 1 + ( 3 4 ) (w) 3 1 = ± 4 w w w = ± (3 4 w) w = 3 4 w 7 4 w = 1 4 w = 1 7 or w = w 7 4 = 1 4 w 7 = w Equation of line : y y 1 = m(x x 1 ) y ( 1) = 1 (x 1) 7 y + 1 = 1 7 x : 7y + 7 = x + 1 x + 7y + 6 = 0 or y ( 1) = 7(x 1) y + 1 = 7x 7 7x y 8 = 0 Comment: Fine. The Dublin School of Grinds Page 136

139 Question 8 a) y y 1 = m(x x 1 ) y ( 6) = 4 (x 7) 5 y + 6 = 4 28 x : 5y + 30 = 4x x + 5y + 2 = 0 1 x 1 y 1 x 2 y 2 (7, 6 ) ( 3, 2 ) m = y 2 y 1 x 2 x 1 = 2 ( 6) 3 7 = x + 5y + 2 = x + 6y 2 = 0 11y = 0 y = 0 => x = 1 2 => ( 1 2, 0) Sketch: x 1 Length from ( 3, y 1 x 2 2 ) to ( 1 y 2):, 0 = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 2 x x 1 y 2 1 y 1 2): Length from (7, 6) to (, 0 = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 2 = ( 1 2 ( 3)) 2 + (0 2) 2 = ( 1 2 7) 2 + (0 ( 6)) 2 = 41 2 = => ratio : 41 2 => 3: 1 The Dublin School of Grinds Page 137

140 b) x 1 = 0 y 1 = 0 a = 1 a b = 1 b c = 1 Square both sides: Perpendicular length = ax 1 + by 1 + c a 2 + b 2 p = (1 a ) (0) + (1 ) (0) + ( 1) b ( 1 2 a ) + ( 1 2 b ) p = 1 1 a b 2 p 2 1 = 1 a b 2 p 2 ( 1 a b 2) = 1 1 a b 2 = 1 p 2 Comment: This is okay. Watch out for part (b) which was asked a long time ago (when you were a baby). Question 9 Cuts y = 0: mx 0 + 4m 2 = 0 mx = 2 4m x = 2 4m m => 2 4m m = x 1 y y 1 = m(x x 1 ) y ( 2) = m(x ( 4)) y + 2 = mx + 4m mx y + 4m 2 = 0 Told x 1 + y 1 = 3 => 2 4m m + 4m 2 = 3 m: 2 4m + 4m 2 2m = 3m 4m 2 9m + 2 = 0 (4m 1)(m 2) = 0 m = 1 4 Cuts x = 0: m(0) y + 4m 2 = 0 y = 4m 2 m = 2 => 4m 2 = y 1 Angle between two lines: tan θ = ± m 1 m m 1 m 2 tan θ = ± (1 4 ) (2) 1 + ( 1 4 ) (2) => tan θ = 7 6 tan θ = ± ( 7 6 ) or => tan θ = 7 6 θ = tan 1 ( 7 6 ) θ = tan 1 ( 7 6 ) θ 49 θ 49 => Acute angle = 49 The Dublin School of Grinds Page 138

141 Question 10 i) Area of a triangle = 1 2 x 1y 2 x 2 y 1 ( 1, 5) (3, 1) ( 2k, 3k) x 1 y 1) x 2 (0, 0) (4, 4 ( 2k + 1, 3k + 5 y 2 ) => 20k + 16 = 56 20k = 40 k = 2 28 = 1 (4)(3k + 5) ( 2k + 1)(4) 2 56 = 12k k 4 56 = 20k + 16 or 20k 16 = 56 20k = 72 k = 16 5 Told k > 0 => k = 2 ii) Call t (x, y) Slope of ts: m = y 2 y 1 x 2 x 1 x 1 y 1) x 2 y 2) t(x, y s(13, 9 => 3 11 = 9 y 13 x x = 99 11y 3x + 11y = Slope of sr: m = y 2 y 1 x 2 x 1 x 1 y 1) x 2 y 2 s(13, 9 r( 1, 5) m = m = 1 => slope of tu = 1 Slope of tu: m = y 2 y 1 x 2 x 1 x 1 y 1 x 2 y 2 t(x, y ) u( 4, 6 ) => 1 = 6 y 4 x 4 x = 6 y x y = x + 11y = x + 3y = 30 14y = 168 => y = 12 => x = 2 => t(2, 12) Comment: Strange yet nice question. Question 11 i) y y 1 = m(x x 1 ) y ( 2) = 3 (x 7) 4 4y + 8 = 3x 21 3x 4y 29 = 0 Slope of K = a b = 3 4 = 3 4 => Slope of M = 3 4 a( 3, 0) p(2, 1) (7, 2) x 1 y 1 => (7, 2) is on M The Dublin School of Grinds Page 139

142 ii) Distance between K and M is twice the perpendicular distance from p to K. x 1 = 2 y 1 = 1 = 2 ( ax 1 + by 1 + c ) a 2 + b 2 a = 3 (3)(2) + ( 4)( 1) + (9) b = 4 = 2 ( ) c = 9 (3) 2 + ( 4) 2 = 2 ( 19 5 ) = 38 5 iii) Slope of ap: Slope of ap: m = y 2 y 1 x 2 x 1 x 1 y 1) x 2, y 2 a( 3, 0 p( 2 1) m = ( 3) m = 1 5 Angle between two lines: tan θ = ± m 1 m m 1 m 2 tan θ = ± (3 4 ) ( 1 5 ) 1 + ( 3 4 ) ( 1 5 ) => tan θ = θ = tan 1 ( ) θ 48 tan θ = ± or => Acute angle = 48 => tan θ = θ = tan 1 ( ) θ 48 iv) ab = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 x 1 y 1) x 2, y 2) a( 3, 0 b(x y 15 = (x ( 3)) 2 + (y 0) 2 15 = (x + 3) 2 + y = x 2 + 6x y 2 x 2 + 6x + y 2 = Also the line K is given by 3x 4y + 9 = 0 2 Rewriting 2 3x = 4y 9 x = 4y 9 3 Subbing into 1 ( 4y 9 2 ) + 6 ( 4y 9 ) + y 2 = : 16y 2 72y y y 2 = y 2 = 2025 y 2 = 81 y = ±9 => x = 4(9) 9 or x = 4( 9) x = 9 x = 15 Told x > 0 => x = 9 (9,9) Comment: Not the easiest question but it s okay. The Dublin School of Grinds Page 140

143 Question 12 i) ii) x 1 = 1 y 1 = 5 a = 3 b = 4 c = 2 x 1 = 1 y 1 = 5 a = 3 b = 4 c = k d = ax 1 + by 1 + c a 2 + b 2 (3)( 1) + ( 4)( 5) + (2) d = (3) 2 + ( 4) 2 d = 15 5 d = 3 (3)( 1) + ( 4)( 5) + (k) 3 = (3) 2 + ( 4) k 3 = 5 15 = 17 + k 15 = 17 + k 2 = k Comment: This is a grand question. or Told k 2 => k = = 17 k k = 32 Question 13 a) L 1 : 3x 2y + 7 = 0 2 L 2 : 10x + 2y x + 13 = 0 x = 1 => y = 2 p( 1, 2) Slope of L 2 = a b = 5 1 = 5 => slope of new line = 1 5 y y 1 = m(x x 1 ) y 2 = 1 (x ( 1)) 5 5y 10 = x + 1 x 5y + 11 = 0 b) (i) y y 1 = m(x x 1 ) y 6 = m(x ( 4)) y 6 = mx + 4m mx y + 4m + 6 = 0 x axis: y = 0 => mx + 4m + 6 = 0 x = 4m 6 m => ( 4m 6, 0) y axis: x = 0 => y + 4m + 6 = 0 y = 4m + 6 => (0, 4m + 6) The Dublin School of Grinds Page 141

144 (iii) Area of a = 1 (base)(perpendicular height) 2 54 = 1 2 (4m + 6 ) (4m + 6) m 108 = 16m2 + 48m + 36 m 108m = 16m m m 2 60m + 36 = 0 4: 4m 2 15m + 9 = 0 (4m 3)(m 3) = 0 m = 3 4 m = 3 Comment: A bit messy. Question 14 (a) Area of = 1 base perpendicular height = 1 2 (base)(10) => base = 25 3 => R: ( 25 3, 0) (b) y y 1 = m(x x 1 ) y 10 = 1.2(x 0) y 10 = 1.2x 1.2x y + 10 = 0 point (0, 10) slope = rise run = 10 ( 25 3 ) = 1.2 Sub ( 5, 4): (c) y = mx + c Sub ( 5, 4): 1.2( 5) (4) + 10 = 0 0 = 0 4 = 5m + c => c = 4 + 5m y = mx x axis: y = y axis: x = 0 => 0 = mx m x = 5m 4 m y = 5m + 4 The Dublin School of Grinds Page 142

145 Area of = 1 base perpendicular height = 1 2 (5m + 4 ) (5m + 4) m m = 250m 3 = m + 25m 2 25m m + 16 = 0 3 m = b ± b2 4ac 2a a = 25 b = c = 16 ( ) ± ( ) 4(25)(16) 2(25) m = 1.2 This is the slope from part (b) or m = 0.53 => c = 4 + 5(0.53 ) = 6. 6 Comment: Very similar to a question from an old The Dublin School of Grinds mock exam. Nothing too difficult. Question 15 (a) Slope of AB: Slope of l 1 Perpendicular: m = y 2 y 1 x 2 x 1 t ( 1) = 7 4 = t = a b = 3 4 = 3 4 => m 1 m 2 = 1 => ( t ) (3 4 ) = 1 t + 1 = 4 t = 5 The Dublin School of Grinds Page 143

146 (b) x 1 = 10 y 1 = k a = 3 b = 4 c = 12 d = ax 1 + by 1 + c a 2 + b 2 (3)(10) + ( 4)(k) + ( 12) d = (3) 2 + ( 4) k d = 5 (c) (i) P must have an equal perpendicular distance l 1 and l 2 : (ii) x 1 = 10 y 1 = k a = 5 b = 12 c = 20 Square both sides: a = 896 b = c = d = ax 1 + by 1 + c a 2 + b 2 (5)(10) + (12)(k) + ( 20) d = (5) 2 + (12) k d = k k => = k + 16k k + 144k2 = k k 2 = k k 2 896k k = 0 k = b ± b2 4ac 2a k = (42336) ± (42336)2 4(896)( 32256) 2(896) ± 4360 k = 1792 k = 3 4 or k = 48 d = 18 4 (3 4 ) 5 d = 3 Comment: Nice question. The Dublin School of Grinds Page 144

147 Co-ordinate Geometry of the Circle Co-ordinate geometry of the circle is worth 4% to 9% of the Leaving Cert. It appears on Paper 2. 1) Equation of a circle with centre (0, 0) and radius r Formula: x 2 + y 2 = r 2 2) Equation of a circle with centre (h, k) and radius r Formula: (x h) 2 + (y k) 2 = r 2 NOTE: This formula appears on Page 19 of The Formulae & Tables Booklet 3) Equation in the form x 2 + y 2 + 2gx + 2fy + c = 0 An equation of a circle in this form has centre ( g, f) and radius g 2 + f 2 c NOTE: This formula and information appears on Page 19 of The Formulae & Tables Booklet NOTE: To represent a circle: i) The coefficients of x 2 and y 2 must be equal ii) The highest power must be 2 iii) There must be no xy term 4) Tangent to a circle We can find the perpendicular distance from the centre of a circle to a tangent using the perpendicular distance formula from Co-Ordinate Geometry of the Line. This distance clearly equals the radius. i.e. radius = ax 1+by 1 +c a 2 +b 2 The Dublin School of Grinds Page 145

148 5) Point(s) of intersection of a circle with an axis When a circle cuts the x-axis: y = 0 When a circle cuts the y-axis: x = 0 There are 3 possibilities: Two points of intersection One point of intersection (i.e. the axis is a tangent to the circle) No points of intersection 6) Point(s) of intersection of a circle and a line RULE: Use simultaneous equations (These are covered in the Algebra chapter) There are 3 possibilities: Two points of intersection One point of intersection (i.e. the line is a tangent to the circle) No points of intersection 7) Points inside/outside/on a circle Remember there are 3 ways for writing the equation of a circle: x 2 + y 2 = r 2 (x h) 2 + (y k) 2 = r 2 x 2 + y 2 + 2gx + 2fy + c = 0 If a point is inside a circle, the left hand side will be < the right hand side. If a point is outside a circle, the left hand side will be > the right hand side If a point is on a circle, the left hand side will be = the right hand side The Dublin School of Grinds Page 146

149 8) Locus A locus is the path traced out by a moving point, satisfying certain given conditions. Example 1 If A(1, 7), B(2, 4) and C(x, y) are three points and AC = 3 BC, show that the locus of C is a circle. Solution: (x 2 x 1 ) 2 + (y 2 y 1 ) 2 = 3 (x 2 x 1 ) 2 + (y 2 y 1 ) 2 A(1, 7) C(x, y) B(2, 4) C(x, y) Square both sides: This is in the form of a circle. (x 1) 2 + (y 7) 2 = 3 (x 2) 2 + (y 4) 2 (x 1) 2 + (y 7) 2 = 9[(x 2) 2 + (y 4) 2 ] x 2 2x y 2 14y + 49 = 9[x 2 4x y 2 8y + 16] x 2 2x y 2 14y + 49 = 9x 2 36x y 2 72y x 2 + 8y 2 34x 58y = 0 8: x 2 + y x y = 0 The Dublin School of Grinds Page 147

150 9) The equation of a tangent In Leaving Cert Higher Level you can be asked to find the equation of a tangent in different scenarios: Scenario (i) Given a point on a circle RULE: Step 1: Find the centre of the circle. Step 2: Find the slope of the radius that contains the point of contact. Step 3: Since the tangent is perpendicular to the radius we can find its slope (by turning the slope upside down and changing the sign). Step 4: Find the equation of the tangent using: y y 1 = m(x x 1 ) This is best explained with an example Example 1 Find the equation of the tangent to the circle x 2 + y 2 + 4x 2y 84 = 0 at the point (3, 7). Solution: Step 1: The equation of the circle is given in the form x 2 + y 2 + 2gx + 2fy + c = 0 so its centre is ( g, f) = ( 2, 1) Step 2: The point of contact is (3, 7) so we can find the slope of the radius using m = y 2 y 1 x 2 x 1 x 1 y 1 x 2 y 2 ( 2, 1) (3, 7) m = ( 2) = 8 5 Step 3: The slope of the tangent = 5 (m 8 1 m 2 = 1) Step 4: y y 1 = m(x x 1 ) Slope = 5 8 Point = (3, 7) y ( 7) = 5 (x 3) 8 y + 7 = 5 (x 3) 8 8y + 56 = 5x 15 5x 8y 71 = 0 Note: An alternative way to answer this question is to use the formula at the bottom of page 19 of the Formulae and Tables Booklet. Scenario (ii) Given a point outside a circle RULE: Step 1: Find the centre and radius of the circle. Step 2: Let the slope of the tangent be m Step 3: Find the equation of the tangent using y y 1 = m(x x 1 ) and write the equation in the form ax + by + c = 0 Step 4: Use the perpendicular distance formula from the tangent to the centre of the circle and solve for m. Step 5: Sub the values for m back into the equation from Step 3. The Dublin School of Grinds Page 148

151 This is best explained with an example Example 2 Find the equations of the tangents from (4, 2) to the circle x 2 + y 2 + 6x 10y + 17 = 0 Solution: Step 1: The equation of the circle is given in the form x 2 + y 2 + 2gx + 2fy + c = 0, so its centre is ( g, f) = ( 3, 5) and its radius is g 2 + f 2 c = ( 3) 2 + (5) 2 17 = 19 Step 2: Let slope of tangent be m Step 3: Equation of tangent: y y 1 = m(x x 1 ) Slope = m Point = (4, 2) y 2 = m(x 4) y 2 = mx 4m mx y + 2 4m = 0 Step 4: We know the distance from the tangent mx y + 2 4m = 0 to the centre ( 3, 5) is the radius length which equals 19 Using the perpendicular distance formula from page 19 of The Formulae & Tables Booklet: Perpendicular distance = ax 1+by 1 +c a 2 +b 2 Square both sides: 19(m 2 + 1) = m + 49m 2 19m = m + 49m 2 30m m 10 = 0 (m)( 3) + ( 1)(5) + (2 4m) 19 = (m) 2 + ( 1) 2 3m m 19 = m m 19 = m m = 3 7m Using the quadratic formula: m = b ± b2 4ac 2a a = 30 b = 42 c = 10 This results in two values for m: m = 0.21 or m = 1.61 Step 5: Equations of the tangents are: mx y + 2 4m = x y + 2 4(0.21) = x y + 2 4( 1.61) = x y = x y = 0 The Dublin School of Grinds Page 149

152 10) Touching Circles The Examiner can ask you to show circles touch externally or internally. If circles touch externally then the distance between their centres equals the sum of their radii: c 1 r 1 r 2 c 2 c 1 c 2 = r 1 + r 2 If circles touch internally then the distance between their centres equals the difference of their radii: c 1 rc 1 c 2 rc 2 c 1 c 2 = r 1 r 2 NOTE: r 1 r 2 has modulus brackets since we may not know which circle is the bigger one and therefore may get a negative answer and obviously a distance can t be negative. The Dublin School of Grinds Page 150

153 11) Problems in g, f and c In some questions we can be given limited information and asked to find the equation of a circle. For these we use simultaneous equations (the ones with three equations, which are explained in the Algebra chapter) Below are three examples: i) Given 3 points: Example 1 Find the equation of the circle containing the points (4, 2), ( 7, 3) and (1, 2) Solution: Let the equation of the circle be: x 2 + y 2 + 2gx + 2fy + c = 0 Sub: (4, 2): (4) 2 + (2) 2 + 2g(4) + 2f(2) + c = 0 8g + 4f + c = 20 1 Sub: ( 7, 3): ( 7) 2 + (3) 2 + 2g( 7) + 2f(3) + c = 0 14g + 6f + c = 58 2 Sub: (1, 2): (1) 2 + ( 2) 2 + 2g(1) + 2f( 2) + c = 0 2g 4f + c = 5 3 Then solve equations 1 2 and 3 using simultaneous equations, which will find g, f and c. Then sub these values back into the original equation. ii) Given 2 points and the equation of the line through the centre: Example 2 Find the equation of the circle which passes through (1, 2) and (3, 2) and whose centre lies on the line 5x + y = 9 Solution: Let the equation of the circle be: x 2 + y 2 + 2gx + 2fy + c = 0 Sub: (1, 2): (1) 2 + (2) 2 + 2g(1) + 2f(2) + c = 0 2g + 4f + c = 5 1 Sub: (3, 2): (3) 2 + ( 2) 2 + 2g(3) + 2f( 2) + c = 0 6g 4f + c = 13 2 Sub: ( g, f) into 5x + y = 9: 5( g) + ( f) = 9 5g f = 9 3 Then solve equations 1 2 and 3 using simultaneous equations, which will find g, f and c. Then sub these values back into the original equation. The Dublin School of Grinds Page 151

154 iii) Given a tangent at a point and one other point: Example 3 Find the equation of the circle through the point (4, 1) which has 2x y 17 = 0 as a tangent at the point (6, 5). Solution: Let the equation of the circle be: x 2 + y 2 + 2gx + 2fy + c = 0 Sub: (4, 1): (4) 2 + (1) 2 + 2g(4) + 2f(1) + c = 0 8g + 2f + c = 17 1 Sub: (6, 5): (6) 2 + ( 5) 2 + 2g(6) + 2f( 5) + c = 0 12g 10f + c = 61 2 The tangent has slope a 2 which equals = 2. Therefore the line perpendicular to this has slope 1. b 1 2 Therefore we can find its equation: y y 1 = m(x x 1 ) y ( 5) = 1 (x 6) 2 y + 5 = 1 (x 6) 2 2y + 10 = x + 6 x + 2y + 4 = 0 But this line also goes through the centre of the circle ( g, f): g + 2( f) + 4 = 0 g 2f = 4 3 Then solve equations 1 2 and 3 using simultaneous equations, which will find g, f and c. Then sub these values back into the original equation. The Dublin School of Grinds Page 152

155 12) Past and probable exam questions Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 The Dublin School of Grinds Page 153

156 Question 8 Question 9 (a) (b) Find the coordinates of P, the point of contact of the tangent 3x 4y + 13 = 0 to the circle x 2 + y 2 + 6y 16 = 0 Find the distance from the centre of the circle to a chord of length 6 of the circle. Question 10 Question 11 (a) (b) c 1 : x 2 + y 2 + 2x 2y 23 = 0 and c 2 : x 2 + y 2 14x 2y + 41 = 0 are two circles. Prove that c 1 and c 2 touch externally. k is a third circle. Both c 1 and c 2 touch k internally. Find the equation of k. Question 12 (a) A circle has centre (-1,5) and passes through the point (1,2). Find the equation of the circle. (b) (i) (ii) The y-axis is a tangent to the circle x 2 + y 2 + 2gx + 2fy + c = 0. Prove that f 2 = c. Find the equations of the circles that pass through the points (-3,6) and (-6,3) and have the y-axis as a tangent. Question 13 (a) Find the centre and radius length of the circle 2x 2 + 2y 2 + 6x 14y 3 = 0. (b) Show that the line 3x 4y + 10 = 0 is a tangent to the circle x 2 + y 2 10x = 0. Question 14 (a) Find the area in terms of π of the following circle: 4x 2 + 4y 2 = 9. (b) Prove that the circles x 2 + y 2 14x 6y + 49 = 0 and x 2 + y 2 6x 12y + 41 = 0 touch externally. The Dublin School of Grinds Page 154

157 Question 15 (a) (b) Question 16 (a) (b) The Dublin School of Grinds Page 155

158 Question 17 Question 18 The Dublin School of Grinds Page 156

159 Question 19 (a) (b) Question 20 Question 21 (a) (b) The Dublin School of Grinds Page 157

160 Question 22 Question 23 (a) (b) Question 24 (a) (b) The Dublin School of Grinds Page 158

161 Question 25 Question 26 Question 27 (a) (b) Question 28 The Dublin School of Grinds Page 159

162 Question 29 Question 30 The Dublin School of Grinds Page 160

163 Question 31 The Dublin School of Grinds Page 161

164 Question 32 The Dublin School of Grinds Page 162

165 13) Solutions to Co-ordinate Geometry of the Circle Question 1 (a) (x h) 2 + (y k) 2 = r 2 (x + 3) 2 + (y 2) 2 = 16 (b) If it s a tangent, then the distance from the centre of the circle to the line equals the radius: radius = g 2 + f 2 c 2g = 2 2f = 4 g = 1, f = 2, c = 15 r = = 20 a = m b = 2 c = 7 x 1 = 1 y 1 = 2 Comment: Fine question. Question 2 centre = ( g, f) = (1, 2) d = ax 1 + by 1 + c a 2 + b 2 m = m m = m 11 20(m 2 + 4) = m 2 22m m m 41 = 0 Use quadratic formula: => m = 41 or m = 1 19 Find P and Q using simultaneous equations. Find midpoint of PQ as centre. Find 1 PQ as radius. 2 Sub into (x h) 2 + (y k) 2 = r 2. x = 20 3y => (20 3y) 2 + y 2 6(20 3y) 8y = y + 9y 2 + y y 8y = 0 10y 2 110y = 0 y 2 11y + 28 = 0 (y 7)(y 4) = 0 y = 7 y = 4 => x = 1 => x = 8 x 1 y 1) x 2, y 2) ( 1, 7 ( 8 4 midpoint = ( 7 2, 11 2 ) radius = 1 2 (x 2 x 1 ) 2 + (y 2 y 1 ) 2 = 1 2 (8 ( 1))2 + (4 7) 2 Comment: No worries. = = => equation: (x ) + (y ) = 45 2 The Dublin School of Grinds Page 163

166 Question 3 To find a where the line is, let s find where it crosses the axis by subbing in x=0 and y=0 Case 1 ( b, b) is on x + 2y 6 = 0 => b + 2b 6 = 0 b = 6 => centre ( 6, 6) radius = 6 (x + 6) 2 + (y 6) 2 = 36 Case 2 (a, a) is on x + 2y 6 = 0 => a + 2a 6 = 0 3a = 6 a = 2 => centre (2, 2) radius = 2 (x 2) 2 + (y 2) 2 = 4 Comment: A bit weird but nothing you can t handle. Know this. Question 4 (a) To find a where the line is, let s find where it crosses the axis by subbing in x=0 and y=0 x = 0 y = 1 2 y = 0 x = 1 radius = 3 centre (x, 3) this is on x 2y 1 = 0 => x 6 1 = 0 x = 7 => (7, 3) => (x 7) 2 + (y 3) 2 = 9 (b) If the touch externally c 1 c 2 = r 1 + r 2 c 1 = (7, 3) r 1 = 3 c 2 = (3, 6) r 2 = ( 3) 2 + ( 6) 2 41 = 2 Comment: A nice question. (3 7) 2 + (6 3) 2 = = 5 5 = 5 The Dublin School of Grinds Page 164

167 Question 5 (a) (b) Midpoint is centre of a circle: x 1, y 1) x 2, y 2 ( 0 7 ( 8 11) PQ = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 PQ = (8 0) 2 + (11 7) 2 PQ = = 80 => radius = ( x 1 + x 2, y 1 + y 2 ) = (4, 9) 2 2 => (x 4) 2 + (y 9) 2 = 20 Slope from centre of circle to Q = y 2 y 1 x 2 x 1 = = 1 2 => Slope of tangent = 2 y y 1 = m(x x 1 ) y 11 = 2(x 8) y 11 = 2x x + y 27 = 0 x-axis y = 0 2x 27 = 0 2x = 27 x = 27 2 => R: ( 27 2, 0) Comment: Very handy. Question 6 Equation of circle: also ( g, f) is on 2x 2y 9 = 0 y = 2x + 2 x 2 + (2x + 2) 2 x (2x + 2) 2 = 0 x 2 + 4x 2 + 8x + 4 x 2x 2 2 = 0 5x 2 + 5x = 0 5x(x + 1) = 0 x = 0 => y = 2 (0,2) or x = 1 => y = 0 ( 1, 0) x 2 + y 2 + 2gx + 2fy + c = 0 sub (0, 2) 4 + 4f + c = 0 1 sub ( 1, 0) 1 2g + c = 0 2 => 2g + 2f 9 = 0 3 solving 123 we get f = 1 g = 7 and c = 8 2 Therefore x 2 + y 2 + 2gx + 2fy + c = 0 becomes x 2 + y 2 7x + 2y 8 = 0 Comment: Standard question. The Dublin School of Grinds Page 165

168 Question 7 (a) x 2 + y 2 5x = 0 2g = 5 centre = ( g, f) => g = 5 = ( 5 2 2, 0) (b) (c) P (4, 2) centre ( 5 2, 0) radius = g 2 + f 2 c = Q (1, 2) RPQ = 5 2 => area of circle = πr 2 = 25 4 π R or S (z, 0) (z 1, 2) x 1 y 1 P (4, 2) (3, 4) x 2 y 2 Q (1, 2) (0, 0) 50π 4 area = 1 2 x 1y 2 x 2 y 2 25π 4 = 1 (z 1)(4) (3)(2) 2 25π 4 = 1 4z π = 4z 10 4 = 4z 10 => z = 50π π 4 x 1 x 1 ( 50π y 1), 4, 50π 0 ( y 2), 0 4 RS = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 = 4z + 10 => z = 50π = ( 50π π ) 2 + (0) 2 Solution continued on the next page. The Dublin School of Grinds Page 166

169 = ( 100π 16 ) 2 = 10,000π2 256 = 100π 16 = 25 4 π Comment: Not nice and very messy but nothing you don t know. Question 8 Circle: centre = (4, 2) radius = g 2 + f 2 c = (4) 2 + ( 2) 2 ( 5) = 5 Radius of k is half of c, ie: 2.5 To find the centre of k we will get the midpoint of the line from c to the intersection of the circles and the tangent: line: x = 4y y 2 8 ( 4y 5 ) + 4y 5 = 0 => ( 4y 5 2 ) 3 3 etc. y = 2 => x = 1 (1, 2) => midpoint = ( , ) 2 = ( 5 2, 0) => k: (x 5 2 ) 2 + (y 0) 2 = (x 5 2 ) 2 + y 2 = 6.25 Comment: Tough question which could separate the top students from the rest. The Dublin School of Grinds Page 167

170 Question 9 (a) 4y 13 x = 3 2 4y 13 ( ) + y 2 + 6y 16 = y 2 104y y 2 + 6y 16 = 0 9 9: 16y 2 104y y y 144 = 0 25y 2 50y + 25 = 0 y 2 2y + 1 = 0 (y 1)(y 1) = 0 y = 1 => x = 3 ( 3, 1) (b) Centre: ( g, f) = (0, 3) Radius: g 2 + f 2 c = = x 2 => x = 4 Comment: Nice short question. Part (b) is an old favourite of the Examiner. The Dublin School of Grinds Page 168

171 Question 10 (a) 2g = 4 2f = 2 => g = 2 f = 1 centre: ( g, f) = (2, 1) radius: g 2 + f 2 c = = 5 (b) So P ( 1, 3) Look at the line from the centre to P. ( 1) 2 + (3) 2 4( 1) + 2(3) 20 = = 0 0 = 0 x 1, y 1 x 2 y 2) ( 2 1) ( 1, 3 m = = 4 3 Comment: Good question. Nothing too hard here. => slope of tangent = 3 4 y y 1 = m(x x 1 ) y 3 = 3 (x + 1) 4 4y 12 = 3x + x axis y = 0 => x = 5 ( 5, 0) The Dublin School of Grinds Page 169

172 Question 11 (a) If touching externally c 1 c 2 = r 1 + r 2 c 1 = ( 1, 1) r 1 = (1) 2 + ( 1) 2 ( 23) = 5 (b) c 2 = (7, 1) r 2 = ( 7) 2 + ( 1) 2 (41) = 3 (7 + 1) 2 + (1 1) 2 = = 8 8 = 8 centre of k: (2, 1) radius = 8 (x h) 2 + (y k) 2 = r 2 (x 2) 2 + (y 1) 2 = 64 Comment: First part is fine but part (b) could be tricky. Watch this! The Dublin School of Grinds Page 170

173 Question 12 (a) radius = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 = = 13 (x + 1) 2 + (y 5) 2 = 13 (b) (i) square both side g = radius g = g 2 + f 2 c g 2 = g 2 + f 2 c => f 2 = c QED (ii) Use x 2 + y 2 + 2gx + 2fy + c = 0 sub ( 3, 6) g + 12f + c = 0 6g + 12f + c + 45 = 0 1 sub ( 6, 3) g + 6f + 5 = 0 12g + 6f + c + 45 = 0 2 But c = f 2 => 1 6g + 12f + f = g + 6f + f = 0 Comment: Nothing surprising here. Know it g 24f 2f 2 90 = g + 6f + f = 0 18f f 2 45 = 0 f f + 45 = 0 (f + 3)(f + 15) = 0 f = 3 => c = 9 => g = 3 x 2 + y 2 + 6x 6y + 9 = 0 f = 15 => c = 225 => g = 15 x 2 + y x 30y = 0 The Dublin School of Grinds Page 171

174 Question 13 (a) 2: x 2 + y 2 + 3x 7y 3 2 = 0 centre: ( g, f) = ( 3 2, 7 2 ) radius = g 2 + f 2 c = ( ) + ( ) ( 3 2 ) (b) If it s a tangent, then the distance from the centre of the circle to the line equals the radius: centre: ( g, f) = (5, 0) radius = g 2 + f 2 c = ( 5) 2 + (0) 2 (0) = 5 = 4 a = 3 b = 4 c = 10 x 1 = 5 y 1 = 0 d = ax 1 + by 1 + c a 2 + b 2 (3)(5) + ( 4)(0) + (10) d = (3) 2 + ( 4) 2 = 25 5 = 5 Comment: Easy. Question 14 (a) 4x 2 + 4y 2 = 9 4: x 2 + y 2 = 9 4 centre: (0, 0) r 2 = 9 4 => r = 3 2 Area = πr 2 = π ( 3 2 ) 2 = 9 4 π (b) If Touching externally: c 1 c 2 = r 1 + r 2 c 1 (7, 3) r 1 = ( 7) 2 + ( 3) 2 (49) = 3 Comment: Standard question with no surprises. c 2 (7, 3) r 2 = ( 3) 2 + ( 6) 2 (41) = 2 (x 2 x 1 ) 2 + (y 2 y 1 ) 2 = ( 4) 2 + (3) 2 = 5 5 = 5 The Dublin School of Grinds Page 172

175 Question 15 a) Use x 2 + y 2 + 2gx + 2fy + c = 0. Sub (1, 1) (1) 2 + ( 1) 2 + 2g(1) + 2f( 1) + c = 0 => 2g 2f + c = 2 1 b) (i) (ii) (iii) Sub ( 6, 2) ( 6) 2 + ( 2) 2 + 2g( 6) + 2f( 2) + c = 0 => 12g 4f + c = 40 2 Sub (3, 5) (3) 2 + ( 5) 2 + 2g(3) + 2f( 5) + c = 0 => 6g 10f + c = 34 3 Now solve 1, 2 and 3 and sub back into x 2 + y 2 + 2gx + 2fy + c = 0 to get: x 2 + y 2 + 4x + 10y + 4 = 0 S 1 : x 2 + y 2 6x 4y + 12 = 0 p(3, 2), r 1 = = 1 = 1 S 2 : x 2 + y y + 20 = 0 q( 5, 2), r 1 = = 9 = 3 L: y 1 = 0 => y = 1 S 1 : x 2 + (1) 2 6x 4(1) + 12 = 0 x 2 6x + 9 = 0 (x 3)(x 3) = 0 => x = 3 As there is only one point (3, 1), L is a tangent to S 1. M: 4x 3y 1 = 0, (x 1, y 1 ) = (3, 2) 4(3) 3(2) 1 d = ( 3) 2 = 5 25 = 5 5 = 1 The perpendicular distance from the centre of S 1 to M is equal to its radius. Therefore, M is a tangent to S 1 Solve L ad M simultaneously: => pw : qw = 5: 3 5- L: y = 1 M: 4x 3y 1 = 0 => 4x 3(1) 1 = 0 4x 4 = 0 4x = 4 x = 1 => w(1, 1) is the point of intersection. pw = (3 1) 2 + (2 1) 2 = = 5 qw = ( 5 1) 2 + ( 2 1) 2 = = 45 = 3 5 = 1: 3 = r 1 : r 2 Comment: An old style question with lots of bits and bobs. Could it make a reappearance?! The Dublin School of Grinds Page 173

176 Question 16 (a) (i) Centre (5k, 3) (ii) Centre (5k, 3), r = 7 => 25k = 7 25k 2 51 = 49 25k 2 = 100 k 2 = 4 => k = 2 as k > 0 (iii) Centre (10, 3), r = 7 Using the perpendicular distance formula: 3(10) + 4( 3) + d 7 = d 7 = 5 35 = 18 + d 18 + d = ±35 (b) Midpoint of [ab] = ( , ) = (4, 2) d = 17 or d = 53 l = (1 4) 2 + (2 ( 2)) 2 = = 5 r 2 = = 125 => r = 5 5 Find the equation of the circles using: x 2 + y 2 + 2gx + 2fy + c = 0 a(1,2) C => g + 4f + c = 0 => 2g + 4f + c = 5 (1) b(7, 6) C => g 12f + c = 0 => 14g 12f + c = 85 (2) The Dublin School of Grinds Page 174

177 The perpendicular distance from the centre of the circle to the line L equals 10. First, find the equation of L. Find the slope of ab. m = The equation of L: Slope = 4 3 Point: (1, 2) = 4 3 y y 1 = m(x x 1 ) y 2 = 4 (x 1) 3 => 4x + 3y 10 = 0 Using the perpendicular distance formula with L and the point ( g, f) 4g 3f = = 4g + 3f g + 3f + 10 = ±50 => 4g + 3f = 40 (3a) or 4g + 3f = 60 (3b) Combine equations (1) and (2) to get: 3g + 4f = 20 (4) Next solve equation (4) with (3a) and (3b) separately to find the equations of C 1 and C 2. Solving equations (4) and (3a) gives Solving equations (4) and (3b) gives f = 8 and g = 4 f = 4 and g = 12 Substituting these values back into equation (1) gives c = 45 => Equation of C 1 : x 2 + y 2 + 8x + 16y 45 = 0 Substituting these values back into equation (1) gives c = 35 => Equation of C 2 : x 2 + y 2 24x 8y + 35 = 0 Comment: Wow! This question would have you going around in circles. Very difficult, yet beautiful all the same. A1 students watch out. Question 17 (a) Slope of op: Slope of Tangent: m = = 9 7 m = 7 9 (b) x 2 + y 2 6x + 4y 12 = 0 Centre (3, 2) r 1 = = 5 x 2 + y x 20y + k = 0 Centre ( 6, 10) r 2 = k r 2 = 136 k p 1 p 2 = r 1 + r 2 p 1 = (3, 2), r 1 = 5 and p 2 = ( 6, 10), r 2 = 136 k ( 6 3) 2 + (10 ( 2)) 2 = k = k 15 = k 10 = 136 k 100 = 136 k => k = 36 The Dublin School of Grinds Page 175

178 (c) Midpoint of [ab]: m = ( 3 + 5, ) m = (1, 2) Find the equation of the circles using: x 2 + y 2 + 2gx + 2fy + c = 0 a( 3, 0) C => 9 6g + c = 0 => 6g + c = 9 (1) b(5, 4) C => g 8f + c = 0 => 10g 8f + c = 41 (2) Let the line through a and b be L. The perpendicular distance from the centre of the circle to the line L equals 5. First, find the equation of L. Find the slope of ab. m = ( 3) The equation of L: Slope = 1 2 Point: (-3, 0) = 1 2 y y 1 = m(x x 1 ) y 0 = 1 (x + 3) 2 => x + 2y + 3 = 0 Using the perpendicular distance formula with L and the point ( g, f) g 2f = g 2f = 5 5 = g + 2f 3 g + 2f 3 = ±5 => g + 2f = 8 (3a) or g + 2f = 2 (3b) Combine equations (1) and (2) to get: 16g 8f = 32 (4) Next solve equation (4) with (3a) and (3b) separately to find the equations of C 1 and C 2. Solving equations (4) and (3a) gives Solving equations (4) and (3b) gives f = 4 and g = 0 f = 0 and g = 2 Substituting these values back into equation (1) gives c = 9 => Equation of C 1 : x 2 + y 2 + 8y 9 = 0 Substituting these values back into equation (1) gives c = 21 => Equation of C 2 : x 2 + y 2 4x 21 = 0 Comment: The last part was tricky. The Dublin School of Grinds Page 176

179 Question 18 (a) r = ( 3 5) 2 + (7 + 8) 2 = = 289 Centre ( 3, 7) Equation of circle: (x + 3) 2 + (y 7) 2 = 289 Or multiplying out you get: x 2 + y 2 + 6x 14y 231 = 0 (b) (i) L: x 3y + 25 = 0 => x = 3y 25 Equation of C: (x + 1) 2 + (y 8) 2 = 160 Substituting x = 3y 25 into the equation of the circle: (3y ) 2 + (y 8) 2 = 160 (3y 24) 2 + (y 8) 2 = 160 9y 2 72y 72y y 2 8y 8y + 64 = y 2 160y = 0 y 2 16y + 48 = 0 (y 12)(y 4) = 0 y = 12 or y = 4 Substitution the y values back into x = 3y 25: x = 11 or x = 13 => p = (11, 12) and q = ( 13, 4) (ii) pq is the diameter if pq is equal to the diameter. pq = ( 13 11) 2 + (4 12) 2 Radius of the circle: = = 640 = 8 10 r = 160 = 4 10 diameter = 2r = 8 10 => [pq]is the diameter. (c) Using the equation of a circle x 2 + y 2 + 2gx + 2fy + c = 0. Substituting (3, 3) into the equation gives: 6g + 6f + c = 18 Substituting (4, 1) into the equation gives: 8g + 2f + c = 17 Use the tangent to find a third equation: Solve equations 1, 2 and 3 simultaneously to find 1 2 Slope of T = 3 slope of pq: 1 3 = 3 + f 3 + g 9 + 3f = 3 g g + 3f = 12 3 g = 9 2, f = 5 2 (ii) The point q is translated through p and on to q. Ans: (6, 2) and c = 24 (3, 3) ( 9 2, 5 2 ) (6,2) Comment: Starts off fine but gets tough towards the end. Watch out for this. The Dublin School of Grinds Page 177

180 Question 19 (a) (i) (ii). (b) (i) (ii) Comment: Good question. The Dublin School of Grinds Page 178

181 Question 20 (a) x 2 + y 2 = r 2 (b) (i) (ii) (c) Circle: x 2 + y 2 12x + 6y + 9 = 0 (ii) ax + by = 0 Substitute this value of x into the circle equation: x = b a y = 4 3 y ( y) + y 2 12 ( 4 y) + 6y + 9 = y 2 90y + 81 = 0 (5y 9)(5y 9) = 0 => y = 9 5 Substitute y = 9 5 back into x = 4 3 y: Ans: ( 12 5, 9 5 ) => x = 12 5 Comment: This is a tough question. In saying that, there s nothing we can t handle here. The Dublin School of Grinds Page 179

182 Question 21 (a) (b) Using the equation of a circle x 2 + y 2 + 2gx + 2fy + c = 0. Substituting (7, 2) into the equation gives: 14g + 4f + c = 53 1 Substituting (7, 10) into the equation gives: 14g + 20f + c = From observation, it can be seen that the 3 rd point is ( 1, 6). Substituting ( 1, 6) into the equation gives: 2g + 12f + c = 37 3 Solving (1), (2) and (3) we get g = 4, f = 6 and c = 27 ( 1, 6) Substituting back into the equation of a circle x 2 + y 2 + gx + 2fy + c = 0: x 2 + y 2 8x 12y + 27 = 0 Comment: Not the nicest question but a clever one. Question 22 (a) (b) (i) Use the perpendicular distance formula: d = ax 1 + by 1 + c a 2 + b 2 (ii) 5 + p( 1) = p p 2 = 6 p 36(1 2 + p 2 ) = 36 12p + p 2 (SBS) 35p p = 0 p(35p + 12) = 0 => p = 0 or p = (c) (i) => K does not intersect S The Dublin School of Grinds Page 180

183 (ii) Start by finding the equation of the line L: Point: ( 2, 2). Slope of K = 4 3 => Slope of L = 3 (perpendicular to K). 4 y y 1 = m(x x 1 ) y ( 2) = 3 (x ( 2)) 4 4y + 8 = 3x + 6 => 3x 4y 2 = 0 Find where L intersects with the circle S by solving simultaneously. Rearranging the equation for L gives: x = 4y Substitute x = 4y into the equation of S: ( 4y Multiplying out and tidying up gives: 2 ) + y ( 4y + 2 ) + 4y 17 = 0 3 y 2 + 4y 5 = 0 (y + 5)(y 1) = 0 => y = 5 or y = 1 Substitute the values for y back into x = 4y + 2 : 3 => x = 6 or x = 2 Therefore, the points of intersection are: ( 6, 5) and (2, 1). The answer is (2, 1) as it is closer to the line. You can check by using the perpendicular distance formula. Comment: An old style question that could make its way back into Project Maths. The Dublin School of Grinds Page 181

184 Question 23 (a) (i) (ii) As can be seen from the diagram, the centre of C 1 lies on C 2 because its radius is twice that of C 2. The point (3, 4) is the mid-point of (2, 3) and the point of contact. (2, 3) (3, 4) (4, 5) (4, 5) is the point of contact between two circles. (b) Using Pythagoras Theorem we can find the radius: r 2 = r 2 = 25 => r = 5 Using this we can now find the centre: Centre: (3, 5) Substituting into the formula: (x h) 2 + (y k) 2 = r 2 The equation of the circle is: (x 3) 2 + (y 5) 2 = 25 Comment: A nice question which required a few clever bits here and there. The Dublin School of Grinds Page 182

185 Question 24 (a) r = ( 3 1) 2 + (2 3) 2 = = 17 Substituing into the equation: (x h) 2 + (y k) 2 = r 2 (x ( 3)) 2 + (y (2)) 2 = ( 17) 2 (x + 3) 2 + (y 2) 2 = 17 (b) The circle has a centre (0,0), so the slope of the radius from the centre to (2, 3) is 3. Therefore, the slope of the 2 tangent is 2. Therefore, the equation of the tangent is: 3 y y 1 = m (x x 1 ) y 3 = 2 (x 2) 3 2x + 3y = 13 To find out where it cuts the x-axis, put y = 0. 2x + 0 = 13 => x = 13 2 The x intercept is ( 13 2, 0) => k = 13 2 Comment: Fine question. Question 25 (a) x 2 + y 2 = ( 2t t 2) + ( 1 t2 1 + t 2) 4t 2 = (1 + t 2 ) 2 + (1 t2 ) 2 (1 + t 2 ) 2 = 4t2 + (1 t 2 ) 2 (1 + t 2 ) 2 = 4t t 2 + t 4 (1 + t 2 ) 2 = 1 + 2t2 + t 4 (1 + t 2 ) 2 = (1 + t2 ) 2 (1 + t 2 ) 2 = 1 (b) (i) Circle: x 2 + y 2 = 10, Centre (0, 0), r = 10 Slope between centre and point of contact: m = 1 3 Perpendicular slope of tangent t: m = 3 Equation of t: Point: (3, 1) Slope: m = 3 y y 1 = m(x x 1 ) y 1 = 3(x 3) y 1 = 3x + 9 => 3x + y 10 = 0 2 The Dublin School of Grinds Page 183

186 (ii) (x 3) 2 + (y + 4) 2 = 50 r = 50 = 5 2 Centre (3, 4) Perpendicular distance to the tangent is equal to the radius. d = ax 1 + by 1 + c a 2 + b 2 1(3) + ( 1)( 4) + k 5 2 = ( 1) k 5 2 = 2 10 = 7 + k 7 + k = ± k = 10 or k = k = 10 k = 17 (c) Common chord is perpendicular to line joining the centres. Call r the midpoint of pq. Midpoint of pq: r(0, 4). Call t the centre of the circle: t( g, f) Slope or pr = = 4 2 = 2 Slope of rt: Perpendicular slope of rt = 1 2 f 4 g 0 = 1 2 f + 4 g = 1 2 => 2f + 8 = g 1 rt = 20 pr = (2 0) 2 + (0 4) 2 = = 20 Distance pr: pt 2 = pr 2 + rt 2 = = 40 pt = 40 ( g 0) 2 + ( f 4) 2 = 20 ( g 0) 2 + ( f 4) 2 = 20 g 2 + f 2 + 8f + 16 = 20 g 2 + f 2 + 8f = 4 2 The Dublin School of Grinds Page 184

187 Replace g in equation 2 by its value from equation 1: (2f + 8) 2 + f 2 + 8f = 4 5f f + 60 = 0 f 2 + 8f + 12 = 0 (f + 6)(f + 2) = 0 f = 6 g = 2f + 8 = 2( 6) + 8 = = 4 Centre: ( g, f) => (4, 6) is a centre or f = 2 g = 2f + 8 = 2( 2) + 8 = = 4 Centre: ( g, f) => ( 4, 2) is a centre Circle 1: Centre (4, 6), r = 40 (x 4) 2 + (y 6) 2 = 40 Circle 2: Centre ( 4, 2), r = 40 (x + 4) 2 + (y 2) 2 = 40 Comment: Bonkers! Very tough question, which shows how the difficulty of question asked has decreased in recent years. A1 students beware of this. Question 26 (a) r = (7 3) 2 + ( 3 + 4) 2 = = 17 Substituing into the equation: (x h) 2 + (y k) 2 = r 2 (x (3)) 2 + (y ( 4)) 2 = ( 17) 2 (x 3) 2 + (y + 4) 2 = 17 (b) (i) x 2 + y 2 8x 10y + 32 = 0 g = 4, f = 5, c = 32 Centre: ( g, f) = (4, 5) r = ( 4) 2 + ( 5) 2 32 = = 9 = 3 (ii) Perpendicular distance to the tangent is equal to the radius. d = ax 1 + by 1 + c a 2 + b 2 3(4) + (4)(5) + k 3 = k 3 = = 32 + k 32 + k = ± k = 15 or k = 17 Comment: Nice question k = 15 k = 47 The Dublin School of Grinds Page 185

188 Question 27 (a) Use x 2 + y 2 + 2gx + 2fy + c = 0. Sub (0, 3) (0) 2 + (3) 2 + 2g(0) + 2f(3) + c = 0 => 6f + c = 9 1 Sub (2, 1) (2) 2 + (1) 2 + 2g(2) + 2f(1) + c = 0 => 4g + 2f + c = 5 2 Sub (6, 5) (6) 2 + (5) 2 + 2g(6) + 2f(5) + c = 0 => 12g + 10f + c = 61 3 Solve equations 1, 2 and 3 simultaneously to get: g = 3, f = 4 and c = 15 and sub back into x 2 + y 2 + 2gx + 2fx + c = 0 to get: x 2 + y 2 6x 8y + 15 = 0 (b) (i) Find the common chord between the circles by subtracting their equations: c 1 : x 2 + y 2 8x + 2y 23 = 0 c 2 : x 2 + y 2 + 6x + 4y + 3 = 0 14x 2y 26 = 0 (equation of the common chord) => 7x + y + 13 = 0 => y = 7x 13 Substitute this value of y back into the equation of c 1 : x 2 + y 2 8x + 2y 23 = 0 x 2 + ( 7x 13) 2 8x + 2( 7x 13) 23 = 0 50x x = 0 5x x + 12 = 0 (5x + 6)(x + 2) = 0 x = 6 5 y = 7x 13 = 7 ( 6 5 ) 13 = 23 5 Point of intersection 1: ( 6 5, 23 5 ) (ii) Call t 1 and t 2, the tangents to c 1 at these points of intersection. c 1 : x 2 + y 2 8x + 2y 23 = 0 centre ( g, f) = (4, 1) c 2 : x 2 + y 2 + 6x + 4y + 3 = 0 centre ( g, f) = ( 3, 2) or x = 2 y = 7x 13 = 1 Point of intersection 2: ( 2,1) = 7( 2) 13 Equation of tangent t 1 Find the slope between the point ( 2, 1) and the centre c 1 (4, 1). m 1 = 1 ( 1) 2 4 = 2 6 = 1 3 The Dublin School of Grinds Page 186

189 Slope of t 1 is perpendicular to this slope: Equation of t 1 : Slope: 3 Point: ( 2, 1) Is ( 3, 2) on t 1? Substitute the point into the equation of t 1 m 1 = 3 y y 1 = m(x x 1 ) y (1) = 3(x ( 2)) y 1 = 3x + 6 => 3x y + 7 = 0 3( 3) ( 2) = 0 => therefore, tangent t 1 passes through the centre of c 2. Equation of tangent t 2 Find the slope between the point ( 6 5, 23 5 ) and the centre c 1 (4, 1). m 2 = 23 5 ( 1) = = Slope of t 2 is perpendicular to this slope: = 9 13 m 2 = 13 9 Solution continued on the next page. Equation of t 1 : Slope: 13 9 Point: ( 6 5, 23 5 ) y y 1 = m(x x 1 ) y ( 23 5 ) = 13 9 (x ( 6 5 )) Is ( 3, 2) on t 2? Substitute the point into the equation of t 1 y = x y = 65x 78 65x + 45y = 0 => 13x + 9y + 57 = 0 13( 3) + 9( 2) = 0 => therefore, tangent t 2 passes through the centre of c 2. Comment: Part (a) is sound but part (b) isn t the nicest question ever! The Dublin School of Grinds Page 187

190 Question 28 (a) (b) (c) (d) Comment: This was the nicest question in years. The Dublin School of Grinds Page 188

191 Question 29 Comment: Scandalous question. What was the Examiner thinking here. Ridiculous. The Dublin School of Grinds Page 189

192 Question 30 (a) (b) (i) (ii) Comment: Good fair question. The Dublin School of Grinds Page 190

193 Question 31 (a) (i) (ii) (iii) Slope of DE: Equation of DE: x 2 + y 2 + 4x + 6y 19 = 0 In the form: x 2 + y 2 + 2gx + 2fy + c = 0 => 2g = 4 g = 2 2f = 6 f = 3 c = 19 centre = ( g, f) = ( 2, 3) ED = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 = ( 2 3) 2 + ( 3 2) 2 = 5 2 but ED = r h + r k 5 2 = r k => r k = 2 m = y 2 y 1 x 2 x 1 = 2 ( 3) 3 ( 2) = 1 y y 1 = m(x x 1 ) y 2 = 1(x 3) => x y 1 = 0 radius = g 2 + f 2 c = (2) 2 + (3) 2 ( 19) = 32 = 4 2 a = 1 b = 1 c = 1 x 1 = 2 y 1 = 2 d = ax 1 + by 1 + c a 2 + b 2 (1)( 2) + ( 1)(2) + ( 1) d = (1) 2 + ( 1) 2 = 5 2 = ie: half of DE (iv) Midpoint of DE = ( x 1 + x 2 2 = ( = ( 1 2, 1 2 ), ) 2, y 1 + y 2 ) 2 Circle j: => ( 1 2, 1 ) ( 2, 2) 2 = (3, 2) ( 1 2, 9 2 ) (x h) 2 + (y k) 2 = r 2 (x ) + (y ) = ( 2) 2 (x ) + (y ) = 2 The Dublin School of Grinds Page 191

194 (v) (b) (i) (ii) Let But using Pythagoras: AB = x AC = y BC = w Area of s = π ( y 2 ) 2 = πy2 4 Area of t = π ( w 2 ) 2 = πw2 4 x 2 = y 2 + w 2 => Area of u = π(y2 + w 2 ) 4 Area of s + Area of t = πy2 πw2 += 4 4 = π(y2 + w 2 ) 4 The Dublin School of Grinds Page 192

195 (iii) From part (ii) above: 1 2 Area of circle u = 1 2 Area of circle s + 1 Area of circle t 2 ie: [2] + [3] + [4] = ([1] + [2]) + ([4] + [5]) [3] = [1] + [5] Comment: A question with way too much English. Students won t like this. Question 32 (a) Centre: (1, 6) Radius: 6 10 (b) (i) k = ( bx 1 + ax 2, by 1 + ay 2 b + a b + a ) (1)(1) + (2)(7) (1)( 6) + (2)(12) k = (, ) (1) + (2) (1) + (2) k = (5,6) (ii) Centre: (5, 6) Radius is 1 3 of 6 10 = 2 10 => (x h) 2 + (y k) 2 = r 2 (x 5) 2 + (y 6) 2 = 40 (c) To find a common tangent we subtract circles: s: (x 1) 2 + (y + 6) 2 = 360 x 2 2x y y + 36 = 360 x 2 + y 2 2x + 12y 323 = 0 1 Comment: Nothing new here. c: (x 5) 2 + (y 6) 2 = 40 x 2 10x y 2 12y + 36 = 40 x 2 + y 2 10x 12y + 21 = 0 2 => 1 2: 8x + 24y 344 = 0 8: x + 3y 43 = 0 The Dublin School of Grinds Page 193

196 Geometry Geometry is worth 4% to 10% of the Leaving Cert. It appears on Paper 2. 1) Theorems There are 3 Theorems (called Theorem 11, Theorem 12 and Theorem 13): Theorem 11 If three parallel lines cut off equal segments on some transversal line, then they will cut off equal segments on any other transversal. Given: Three parallel lines l, m and n, intersecting the trasversal t at the points A, B and C such that AB = BC. Another transversal, k, intersects the lines at D, E and F D P A l E B m Q F C n k t To Prove: DE = EF Construction: Through E, construct a line parallel to t and intersecting l at the point P and intersecting n at the point Q. Proof: PEBA and EQCB are parallelograms Then PE = AB and EQ = BC. Opposite sides of a parallelogram But AB = BC So, PE = EQ (Side) PED = FEQ Vertically opposite angles (Angle) DPE = FQE Alternate angles (Angle) Angle/Side/Angle DEP and FEQ are congruent DE = EF The Dublin School of Grinds Page 194

197 Theorem 12 Let ABC be a triangle. If a line is parallel to BC and cuts AB in the ratio s: t, then it also cuts AC in the same ratio. Given: A triangle ABC and a line XY parallel to BC which cuts AB in the ratio s: t A X Y To Prove: AY : YC = s: t B C Construction: Divide AX into s equal parts and XB into t equal parts. Through each point of division, draw a line parallel to BC: A k X Y B C Proof: According to Theorem 11, the parallel lines cut off segments of equal length along AC. Let k be the length of each of these equal segments. AY = sk and YC = tk AY YC = sk tk = s t The Dublin School of Grinds Page 195

198 Theorem 13 If two triangles ABC and DEF are similar, then their sides are proportional, in order: AB = BC = AC DE EF DF Given: The triangles ABC and DEF in which 1 = 4, 2 = 5 and 3 = 6 A 1 D 4 B C E F To Prove: AB DE = BC EF = AC DF Construction: Mark the point X on [AB] such that AX = DE. Mark the point Y on [AC] such that AY = DF Join XY: A 1 D X 5 6 Y 4 B 2 3 C E 5 6 F Proof: The triangles AXY and DEF are congruent..side/angle/side AXY = DEF..Corresponding angles AXY = ABC XY BC AB AX = AC AY AB DE = AC DF Similarly, it can be proven that AB = BC DE EF.. A line parallel to one side divides the other side in the same ratio (Theorem 12) AB DE = BC EF = AC DF The Dublin School of Grinds Page 196

199 2) Definitions There are 9 definitions. Learn them along with the examples: 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 term. 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. eg. 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. eg. 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) Implies: To imply something is to use something we have proved previously. The symbol for implies is => or eg. a + b = 15 3 => a + b = 5 6) Proof: A proof is a series of logical steps we use to prove a theorem. eg. (You will only get asked this definition if you are asked to do one of the Theorems. Therefore you will use the theorem as your example for the proof). The Dublin School of Grinds Page 197

200 7) Proof by Contradiction: A proof by contradiction is a proof where we make an assumption, then prove it is false by using valid axioms. Alternatively you can state something is false and prove yourself wrong, meaning it is true. Example: Prove that a triangle cannot have 2 right angles. Assume a triangle CAN have 2 right angles. Now, the sum of the interior angles of a triangle is c = c = 180 c = 0. This implies that the third angle has to be 0. However, if the third angle is 0, this means that all 3 points of the triangle are collinear (in a straight line). Therefore, by contradiction a triangle can't have 2 right angles. 8) Is Equivalent to: This means that something has the same value or measure as, or corresponds to, something else. Eg: 5 2 has the same value as 0.4 9) If and only if: This means that one thing is true (or false) only if another thing is true (or false). Eg: A parallelogram is a rhombus if, and only if, all four of its sides have the same lengths. The Dublin School of Grinds Page 198

201 3) Constructions At Leaving Cert Higher Level there are 22 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 4) Basic rules of angles and triangles We must look at some basic rules of angles and triangles which we would have covered at Junior Cert level. Point 1: There are 180 in a straight angle Point 2: There are 90 in a right angle Point 3: There are 360 in a full circle Point 4: Vertically opposite angles are equal Example 1: A = B C = D Point 5: Alternate angles (or Z angles) are equal. Example 2: A = B C = D Note: These rules are only true when the lines are parallel as shown by Point 6: Corresponding angles (or F angles) are equal Example 3: A = B C = D Point 7: There are 180 in any triangle in the world Point 8: An equilateral triangle, or an equiangular triangle, is a triangle where the sides and angles are equal Example 4: The Dublin School of Grinds Page 200

203 Note: Equal sides are often labelled with little dashes as shown in the diagram above. Sometimes they use two dashes: Point 9: An isosceles triangle is a triangle which has two equal sides. This implies that there are two equal angles also. Example 5: A C A C Point 10: The exterior angle of a triangle equals the two opposite interior angles added together. Example 6: A = C + D The next two rules may be new to you: Point 11: The sum of any two lengths of a triangle must be greater than the other side. Example 7: Example 8: 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 14 We can write this as 6 x > > > 8 The Dublin School of Grinds Page 201

204 Point 12: The largest angle is across from the largest side & The smallest angle is across from the smallest side Example 9: Which side is the largest in the following diagram: Answer: [xy] The Dublin School of Grinds Page 202

205 5) Congruent triangles Congruent triangles are triangles that are the same as each other. They may be flipped upside down and/or left to right but they are still the same. Example 1: ABC is congruent to DEF this can be written as ABC DEF You may have noticed that every triangle in the world has six things : 3 sides and 3 angles. To prove triangles are congruent we don t have to show all six things are equal, we only have to show that one of the following four are true: 1) SSS (Side/Side/Side) 2) SAS (Side/Angle/Side) 3) ASA (Angle/Side/Angle) 4) RHS (Right angle/hypotenuse/side) Note 1: The areas of congruent triangles are equal. Note 2: If we prove three things are equal, then we can say the other three things are equal. The Dublin School of Grinds Page 203

206 6) Quadrilaterals Quadrilaterals are four sided figures. You need to know the properties of quadrilaterals: Type 1: A Square Has four equal sides Opposite sides are parallel All angles are 90 Diagonals bisect each other (Bisect means cuts exactly in half): 90 angle formed Type 2: A Rectangle Opposite sides are equal Opposite sides are parallel All angles are 90 Diagonals bisect each other Type 3: A Parallelogram Opposite sides are equal Opposite sides are parallel Opposite angles are equal Diagonals bisect each other 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 204

207 Type 4: A Rhombus (this is a pushed over square) Has four equal sides Opposite sides are parallel Opposite angles are equal Diagonals bisect each other (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 The Dublin School of Grinds Page 205

208 7) Parallel line rules There are two rules we need to learn in this section: 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: For example: 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 For example: 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 top length = bottom length bottom length top length 3 1 = 6 2 ie. 3 = 3 TRUE 1 3 = 2 6 bottom length overall length overall length bottom length top length overall length = overall length top length = bottom length overall length = overall length bottom length = top length overall length overall length top length ie. 1 3 = 1 3 TRUE 1 4 = 2 8 ie. 1 4 = 1 4 TRUE 4 1 = 8 2 ie. 4 = 4 TRUE 3 4 = 6 8 ie. 3 4 = 3 4 TRUE 4 3 = = 4 3 TRUE The Dublin School of Grinds Page 206

209 We must be able to apply these rules to problems: Example 1: 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 207

210 8) Similar triangles Similar triangles are triangles that have equal corresponding angles. The corresponding sides are in the same ratio. This is best illustrated with an example. From this rule we can use many ratios: Example 1: Example 2: small base big base small right = big right 6 12 = 5 10 ie: 1 2 = 1 2 small left = big left 8 16 = 5 10 ie: 1 2 = 1 2 small right big right We could also do ratios within the same triangles: Example 3: small left small bottom = 8 6 = ie. 4 3 = 4 3 big left big bottom As you can see there are many ratios we can use. In previous sections 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 208

211 9) Area of triangles and parallelograms There is a very basic formula we should know from the Junior Cert. Example 1: Area of triangle = 1 (base)(perpendicular height) area = 1 2 (18)(6) = 54 units 2 Also: Area of paralellogram = (base)(perpendicular height) Example 2: area = (10)(3) = 30 units 2 The Dublin School of Grinds Page 209

212 10) 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 vi) Tangent arc 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 210

213 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 211

214 11) Transformations There are 3 types of transformations to look at in this section: 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. Example 1: B The letter F has no axis of symmetry A square has 4 axis of symmetry The letter B has 1 axis of symmetry The Dublin School of Grinds Page 212

215 Example 2: Draw a shape which has exactly 3 axes of symmetry. Show the axes on the diagram. Example 3: Draw a shape which has exactly 2 axes of symmetry. Show the axes on the diagram. iii) Central symmetry Central symmetry reflects a figure through a point. It looks back to front and upside 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 213

216 12) 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. How do I actually find and draw the enlarged image? Important rules: Scale factor (k) = Image length Object length Image area = k 2 Object area The Dublin School of Grinds Page 214

217 13) Past and probable exam questions Question 1 The Dublin School of Grinds Page 215

218 Question 2 Question 3 The Dublin School of Grinds Page 216

219 Question 4 Question 5 The Dublin School of Grinds Page 217

220 Question 6 Question 7 The Dublin School of Grinds Page 218