ANALYTICAL GEOMETRY Revision of Grade 10 Analytical Geometry

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1 ANALYTICAL GEOMETRY Revision of Grade 10 Analtical Geometr Let s quickl have a look at the analtical geometr ou learnt in Grade LESSON Midpoint formula (_ + 1 ;_ + 1 The midpoint formula is used to find the coordinates of the midpoint of a line segment. Find the midpoint of the line joining A( ; and B(4; 6 M B(4; 6 A( ; To find the co-ordinate of the midpoint we add the coordinates of A and B together and divide b _ = 1 We then do the same using the values of A and B _ = 4 The coordinates of the mid-point are therefore M(1; 4. Distance formula The distance formula is used to find the distance between two points A( 1 ; 1 and B( ;. AB = ( + ( 1 Find the distance between A(; and B(7; 4 When using the distance formula we must remember that if we start with point A we must keep on using A first. AB = ( A B + ( A B = ( 7 + ( 4 A B A B = ( 5 + ( 1 _ 6 = 009 Page 11

2 The question might ask ou to leave our answer in simplest surd form or to 1 or decimal places. Gradient (slope The gradient of a line gives us an indication of the steepness of the line. It is important to remember that a line with a positive gradient runs uphill from left to right. A line with a negative gradient runs downhill if we look at it from left to right. When calculating the gradient of a line we must alwas look at the answer and make sure it looks right. m AB 1 Calculate the gradient of the line AB. A( 1; ( 4 m AB = _ B(; 4 Our answer is negative and the line runs downhill from left to right so this looks correct. Let s quickl talk about lines that are parallel. Parallel lines have the same gradient. m 1 = m What about perpendicular lines? Perpendicular lines intersect at an angle of 90. For perpendicular lines m 1 m = 1 D C B F This means if we multipl the gradients with each other our answer must be 1. eg. m BC m DF = _ m BC m DF _ = 1 The answer is 1 so the lines BC and DF must be perpendicular. Page 114

3 Colinear points Do ou still remember what colinear points are? Colinear points are points that lie on the same straight line. A B C The points A, B and C lie on the same straight line and are therefore colinear. This means that m AB = m BC = m AC. To prove that points are colinear we onl have to prove that of the gradients are equal. Now that we have revised what ou learnt in Grade 10 we will use these skills to learn the following: Finding the equation of a line. More about parallel and perpendicular lines. Inclination of a line. After we have gone through these topics we will then do an eercise where we test all of our analtical geometr skills! Finding the equation of a line. The equation of a straight line is given b = m + c m is the gradient of the line and c is the intercept 1 Give the equation of a straight line with gradient = 4 and -intercept. This is eas, all we have to do is substitute m and c = 4 Find the equation of the line with gradient = 4 passing through A(; 8 We use the formula 1 = m( 8 = 4( value of A value of A substitute the gradient 8 = = Find the equation of the straight line joining A( ; and B (1; _ 009 Lesson 1 Algebra Page 115 Page 1

4 First we must calculate the gradient between A and B using the gradient formula m AB 1 = _ 1 _ = _ = _ 1 then we use the formula 1 = m( Into this formula we substitute the gradient and one of our points. I will use A( ; = _ 1 ( ( = _ 1 1 = _ 1 + Special lines 4 Consider the line above: This line is parallel to the ais and intersects the ais at 4. These lines have the general equation = c where c is the intercept. The equation of the line is therefore = 4 The line above is parallel to the ais and intersects the ais at. These lines have the general equation = c where c is the intercept. The equation of the line is therefore =. Page 116

5 4 P( ; 1 Q(; 5 and R(4; are the vertices of PQR. M is the midpoint of QR. Determine the equation of line PM. It is alwas helpful to draw a rough diagram. Q(; 5 P( ; 1 M R(4; We first have to find the coordinates of M. M is the midpoint of the line QR so we use the midpoint formula (_ + 1 ;_ 1 + (_ + 4 ; _ 5 + ( M(; 1 Now that we have the coordinates of M we use the gradient formula to calculate the gradient of PM. m PM 1 This is interesting!! 1 1 What does it mean if 0 our gradient is zero? 5 = 0 It means the line is parallel to the -ais. Therefore the equation is = 1. (Please note that had the gradient not been zero we would have used the formula 1 = m( to calculate the equation in the form = m + c. Parallel and perpendicular lines We revised earlier that parallel lines have the same gradient and perpendicular lines have gradients with product equal to 1. 5 Show that the line AB defined b = + is perpendicular to the line CD, defined b + = 0 Both equations must be in standard form. = + + = 0 = + = _ Lesson 1 Algebra Page 117 Page 1

6 Then compare the gradients m AB = m CD = _ 1 m AB m CD = 1 Therefore the lines AB and CD are perpendicular. 6 Determine the equation of the line AB if it is given that AB is perpendicular to the line = and passes through ( ; 8. Write the equation of the given line in standard form. = = + 4 The line has a gradient of +. This means that the gradient of AB = _ 1 as it is given that the lines are perpendicular. Now we use the formula 1 = m( to determine the equation of the line. ( 8 = _ 1 ( ( + 8 = _ 1 1 = _ 1 9 Perpendicular bisectors A It is important to know the properties of a perpendicular bisector. The name tells us all we need to know! C In the diagram alongside the line AB is the perpendicular bisector of CD. AB is perpendicular to CD and it cuts CD into equal parts. (Passes through the midpoint. D 7 Find the equation of the perpendicular bisector of AB if A( 6; and B(; 1. As alwas it is a good idea to make a rough sketch. B B(; 1 M A( 6; Page 118

7 First we calculate the coordinates of M the midpoint of AB as our perpendicular bisector passes throught the point M. We use the midpoint formula. (_ + 1 ;_ 1 + (_ 6 + ;_ + 1 M (_ ; 1_ Now we calculate the gradient of the line AB. m AB We know that our perpendicular bisector must have a gradient of as m 1 m = 1 for perpendicular lines. We now use the coordinates of the midpoint as well as the gradient to find the equation of the perpendicular bisector. 1 = m( ( _ 1 = [ ( _ ] ( _ 1 = ( + _ + _ 1 = [ + _ ] + _ 1 = _ 9 + _ 1 = _ 9 ( = = = 6 9 = 6 10 = 5 8 A( 10; B(10; 8 and C(11; 9 are the vertices of ABC. Determine the equation of CT If CT is perpendicular to AB. First we determine the gradient of AB as we will T then be able to determine the gradient of CT. m AB 1 8 _ A( 10; \ the gradient of CT =. We now use C(11; 9 and m = to determine the equation of CT. 1 = m( ( 9 = ( = + = + 1 B(10; 8 C(11; Lesson 1 Algebra Page 119 Page 1

8 Angle of Inclination The angle of inclination of a line is the angle that the line makes with the positive -ais. B A Consider the line AB above. AB has a positive gradient as the line runs uphill from left to right. The angle of inclination q is an acute angle. Consider the line CD. CD has a negative gradient as CD runs downhill from left to right. The C angle of inclination q is an obtuse angle. The size of the angle of inclination can be calculated using tan q = m D m tanθ tan θ = m 9 Calculate the angle which the line passing through ( ; and (; 4 makes with the positive ais. (; 4 ( ; First we calculate the gradient of the line. m 1 4 ( ( 6 5 The line has a positive gradient so we know that q will be an acute angle. tan q = m tan q 6 5 We calculate q b using the tan 1 function on our calculator. q = 50, Page 10

9 10 Calculate the angle between the line passing through ( ; and (; 4 and the positive ais. ( ; (; 4 Again we start b calculating the gradient of our line. m 1 ( The line has a negative gradient so we know that q will be an obtuse angle. tan q = m tan q 7 5 We calculate a reference angle b using the tan 1 function on our calculator. Ref angle = 54,5 Now we add 180 to give us an angle between 90 and 180. q = 54, = 15,5 We onl add 180 if our gradient is negative!!!!! Let s have a look at a slightl more difficult eample. 11 P( 4; 5 Q(; 4 R( 1; 4 Calculate the size of angle β. 009 Lesson 1 Algebra Page 11 Page 1

10 In order to calculate the size of angle B we will first have to calculate a and q. m PR 1 m 1 QR 4 ( 4 ( 1 _ 5 ( 4 = 4 ( 1 9 tan q = = q = 6,4 tan a = Ref Ð = 71,6 a = 108,4 Now let s look at the diagram again (supplementar angles 6.4 (verticall opposite Ð B = 180 (71,6 + 6,4 = 45 (angles of a triangle add up to 180 Activit Activit Practise the questions in this section to ensure ou understand all the concepts in Analtical geometr. 1. Determine the equation of the straight line that passes through: 1.1 the points A(; 4 and B(1; 5 1. the point ( 6; and perpendicular to + = 6 1. (; 1 and parallel to 6 = 1. A( 5; 5, B(4; 5 and C(1; p are colinear. Determine.1 the length of AB. the gradient of AB. the angle AB makes with the positive ais.4 the value of p.. Given A( 5; 1, B(1; 6 and C(7;. Determine.1 the length of AC Page 1

11 . the equation of the line BC. the coordinates of P the midpoint of AB.4 the equation of the line parallel to AC passing through the point ( 1; 4. Determine the equation of the line through the point ( ; and perpendicular to the line + = A( 4;, B(6;, T(8; a and S(4; t and points in a cartesian plane. 5.1 Determine the equation of the line AB. 5. Calculate the value of a if BT^AB 5. Calculate the value of b if M(b; 0 is the midpoint of BT _ 5.4 Calculate the values of t if BS = In the figure PQR is a triangle. The coordinates of P are (1; 1. The equations of QR and PR are + 6 = 0 and = 0 respectivel. 6.1 Show that coordinates of Q are (0; 6. Calculate the gradient of QR 6. Prove that P ^ Q R = 90 Q 6.4 Find the coordinates of R 6.5 Calculate the length of RQ P(1; 1 R 7. Calculate the gradient of the line with inclination Find the equation of the perpendicular bisector of MN if M is the point ( ; and N is the point (5; 8 9. ABCD is a parallelogram with vertices A(;, B(7; 5, C(; 11 and D(;. Find the 4 th verte of the paralellogram. 10. Find the obtuse angle between AB and CD IF A( 5; 4, B(5; and the equation of CD is + 10 = 0 A ( 5; 4 D obtuse angle (; (5; 5 B C 009 Lesson 1 Algebra Page 1 Page 1

12 s 1.1 m AB = m( ( 4 = _ 1 ( 8 = + = + 11 = _ 1 + _ = 6 = + 6 = _ + m = _ m of ^ line 1 = m( ( ( ( 6 ( + ( = ( = + 1 = = 1 = = + _ 1 1 = m( ( 1 = ( + 1 = 9 = AB = ( + ( 1 = ( (5 ( 5 = _ = 181. m AB 1 5 ( tan q = m tan q 10 9 Ref q = 48 q = 1 Page 14

13 .4 m BC = m AB _ p p + 45 = 0 p AC = ( + ( 1 = ( (1 ( = _ = 15. First calculate the gradient of BC m BC 1 6 ( = 4 _ 1 = m( 6 = _ 4 ( 1 18 = = 4 + = _ 4 + _. (_ + 1 ;_ 1 + (_ ;_ P ( ; _ 7.4 m AC 1 7 ( = m( 1 ( ( = 1 4 = _ First write equation of line in standard form = _ 1 gradient \ ^ line gradient m 1 = m( ( ( ( + ( = Lesson 1 Algebra Page 15 Page 1

14 = 5.1 m AB ( 4 6 = 5_ 10 = _ 1 1 = m( = _ 1 ( ( 4 6 = 4 = + = _ m AB 1 \ m BT = if AB^BT m BT a ( 8 6 = _ a + 1 a + = 4 a = 5. (_ + 1 ;_ 1 + _ = b b = 14 b = BS = ( + ( 1 BS = (6 4 + ( t = t + t = t + 4t + 8 BS = t + 4t + 8 _ ( 10 = t + 4t = t + 4t (t + 8(t 4 = 0 t = 8 or t = First put both equations into std form. = 0 = + = = 6 Must be RP ( intercept 1 + Must be QR (+ intercept -intercept = ( = 0 \ Q = (0; 6. m QR 1 (see 6.1 Page 16

15 6. m QP 1 Use Q ( = m QR m QP 1 ( = 1 \ P ^ Q R = R is the point of intersection of QR and RP 1_ + = + 6 = 6 = 1 = 6 to find value sub = 6 back into either equation = (6 = 4 R(6; RQ = ( + ( 1 = (6 0 + (4 _ = θ = 100 tan 100 = m m = 5, = + m = 1 = m( + = ( 4 = m PQ 1 ( 7 ( m = P( 1; sub into 1 = m ( + = ( + 1 = = 5 8. m MN 8 5 ( 10 8 = _ 5 4 m 4 5 Midpoint MN (_ + 5 ;_ Lesson 1 Algebra Page 17 Page 1

16 (1; 1 = m(n n 1 Þ ( 4 ( = 4n 4 5 = 4n n _ Midpoint AC (_ + ;_ 11 (_ 5 ; 4 Midpoint BD (_ 7 + ; _ 5 + _ = _ 5 + = = 8 = 10. m AB tan θ 5 θ = 149,04 CD: = 10 5 m CD tan a a = 56,1 between AB and CD = 149,04 56,1 = 9,7 Page 18

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