F4 Add Maths - Coordinate Geometry

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F4 Add Maths - Coordinate Geometry

  1. 1. CHAPTER 6 COORDINATE GEOMETRY FORM 4 20 PAPER 1 1. A point T divides the line segment joining the points A(1, -2) and B(-5, 4) internally in the ratio 2 : 1. Find the coordinates of point T. [2 marks] 2. Diagram below shows a straight line PQ with the equation 3 x + 5 y = 1. The point Q lies on the x-axis and the point P lies on the y-axis. Find the equation of the straight line perpendicular to PQ and passing through the point Q. [3 marks] 3. The line 8x + 4hy - 6 = 0 is perpendicular to the line 3x + y = 16. Find the value of h. [3 marks] 4. Diagram below shows the straight line AB which is perpendicular to the straight line CB at the point B. The equation of the straight line CB is y = 3x  4. Find the coordinates of B. [3 marks] 5. The straight line 14 x + m y = 1 has a y-intercept of 3 and is parallel to the straight line y + nx = 0. Determine the value of m and of n. x P Q y 0 A(0,6) B x y C 0
  2. 2. CHAPTER 6 COORDINATE GEOMETRY FORM 4 21 [3 marks] 6. Diagram below shows a straight line passing through A(2, 0) and B (0, 6). a) Write down the equation of the straight line AB in the form a x + b y = 1. [1 mark] b) A point P(x, y) moves such that PA = PB. Find the equation of the locus of P. [2 marks] x B(0, 6) A(2, 0) y 0
  3. 3. CHAPTER 6 COORDINATE GEOMETRY FORM 4 22 PAPER 2 1. Solutions to this question by scale drawing will not be accepted. Diagram shows a straight line CD which meets a straight line AB at the point D. The point C lies on the y-axis. 0 a) Write down the equation of AB in the form of intercepts. [1 mark ] b) Given that 2AD = DB, find the coordinates of D. [2 marks] c) Given that CD is perpendicular to AB , find the y-intercept of CD. [3 marks] 2. Solutions to this question by scale drawing will not be accepted. In the diagram the straight line BC has an equation of 3y + x + 6 = 0 and is perpendicular to straight line AB at point B. (a) Find i) the equation of the straight line AB ii) the coordinates of B. [5 marks] (b) The straight line AB is extended to a point D such that AB : BD = 2 : 3. Find the coordinates of D. [2 marks] (c) A point P moves such that its distance from point A is always 5 units. Find the equation of the locus of P. [3 marks] 0 x y DA(0 , -3) C B (12, 0) A(-6, 5) B C 3y + x + 6 = 0 x y 0
  4. 4. CHAPTER 6 COORDINATE GEOMETRY FORM 4 23 3. Solutions to this question by scale drawing will not be accepted. Diagram shows the triangle AOB where O is the origin. Point C lies on the straight line AB. (a) Calculate the area, in unit2 , of triangle AOB. [2 marks] (b) Given that AC : CB = 3 : 2, find the coordinates of C. [2 marks] (c) A point P moves such that its distance from point A is always twice its distance from point B. (i) Find the equation of the locus of P. (ii) Hence, determine whether or not this locus intercepts the y-axis. [6 marks] 4. In the diagram, the straight line PQ has an equation of y + 3x + 9 = 0. PQ intersects the x-axis at point P and the y-axis at point Q. Point R lies on PQ such that PR : RQ = 1 : 2. Find (a) the coordinates of R, [3 marks] (b) the equation of the straight line that passes through R and perpendicular to PQ. [3 marks] y + 3x + 9 = 0 y x A(-2, 5) B(5, -1) 0 C y x P Q 0 R
  5. 5. CHAPTER 6 COORDINATE GEOMETRY FORM 4 24 5. Solutions to this question by scale drawing will not be accepted. Diagram shows the triangle OPQ. Point S lies on the line PQ. a) A point W moves such that its distance from point S is always 2 2 1 units. Find the equation of the locus of W. [3 marks] b) It is given that point P and point Q lie on the locus of W. Calculate i) the value of k, ii) the coordinates of Q. [5 marks] c) Hence, find the area , in unit2 , of triangle OPQ. [2 marks] 0 x y P(3 , k) S (5, 1) Q
  6. 6. CHAPTER 6 COORDINATE GEOMETRY FORM 4 25 ANSWERS ( PAPER 1 ) 1. T ( 3 )2)(5()1)(1(  , 3 )2)(4()1)(2(  ) 2 = T( -3 , 2 ) 1 2. Gradient of PQ , m1 = - 3 5 and the coordinates of Q (3 , 0) 1 Let the gradient of straight line perpendicular to PQ and passing through Q = m2 . Then m1  m2 = -1. m2 = 5 3  The equation of straight line is 3 0   x y = 5 3 5y = 3(x – 3) 1 5y = 3x – 9 1 3. Given 8x + 4hy – 6 = 0 4hy = -8x + 6 y = - h4 8 x + h4 6 y = - h 2 x + h2 3 Gradient , m1 = - h 2 3x + y = 16 y = -3x + 16 Gradient , m2 = -3 1 Since the straight lines are perpendicular to each other , then m1  m2 = -1.  (- h 2 )(-3) = -1 1 6 = -h h = -6 1 4. Gradient of CB , m1 = 3 Since AB is perpendicular to CB, then m1 m2 = 1 Gradient of AB, m2 =  3 1 1  The equation of AB is y = - 3 1 x + 6 B is the point of intersection. y = 3x  4 ……………(1) y =  3 1 x + 6 ……………(2) 3x  4 =  3 1 x + 6 1
  7. 7. CHAPTER 6 COORDINATE GEOMETRY FORM 4 26 3 10 x = 10 x = 3 y = 3(3)  4 = 5 The coordinates of B are (3, 5). 1 5. 14 x + m y = 1  y-intercept = m = 3 1 From 14 x + 3 y = 1, the gradient m1 = - 14 3 From y = -nx , the gradient m2 = -n . Since the two straight lines are parallel , then m1 = m2 - 14 3 = -n 1  n = 14 3 1 6. a) From the graph given, x- intercept = 2 and y-intercept = 6. The equation of AB is 2 x + 6 y = 1 . 1 b) Let the coordinates of P = (x , y) and since PA = PB 22 )0()2(  yx = 22 )6()0(  yx (x – 2)2 + y2 = x2 + (y – 6)2 x2 – 4x + 4 + y2 = x2 + y2 – 12y + 36 1 12y – 4x -32 = 0 3y – x - 8 = 0 1
  8. 8. CHAPTER 6 COORDINATE GEOMETRY FORM 4 27 ANSWERS ( PAPER 2 ) 1 a) 12 x - 3 y = 1 1 b) Given 2AD = DB , so DB AD = 2 1  D = ( 3 )1(12)2(0  , 3 )1(0)2(3  ) 1 = ( 4 , -2 ) 1 c) Gradient of AB, mAB = -( 12 3 ) = 4 1 1 Since AB is perpendicular to CD, then mAB mCD = 1.  Gradient of CD, mCD = - 4 Let, coordinates of C = (0 , h) , mCD = 40 )2(  h - 4 = 4 2  h 16 = h + 2 h = 14 1  y-intercept of CD = 14 1 2 a) i) Given equation of BC, 3y + x + 6 = 0 y = - 3 1 x – 2 Gradient of BC = - 3 1 1 Since AB is perpendicular to BC , then mAB mBC = 1. Gradient of AB, mAB = 3 The equation of AB , )6( 5   x y = 3 y – 5 = 3x + 18 1 y = 3x + 23 1 ii) B is the point of intersection. Equation of AB , y = 3x + 23 …………. (1) Equation of BC , 3y + x + 6 = 0 ………….(2) Substitute (1) into (2), 3(3x + 23) + x + 6 = 0 1
  9. 9. CHAPTER 6 COORDINATE GEOMETRY FORM 4 28 9x + 69 + x + 6 = 0 x = - 2 15 Substitute value of x into (1), y = 3(- 2 15 ) + 23 y = 2 1  The coordinates of B are ( - 2 15 , 2 1 ) 1 b) Let D (h, k) B( - 2 15 , 2 1 ) = ( 5 )18(2 h , 5 152 k ) 1 - 2 15 = 5 )18(2 h , -75 = 4h – 36 h = 4 39 2 1 = 5 152 k 5 = 4k + 30 k = 4 25  1 The coordinates of D are ( 4 39 , 4 25  ) c) Given PA = 5 22 )5())6((  yx = 5 1 ( x + 6)2 + ( y – 5)2 = 25 1 x2 + 12x + 36 + y2 -10y + 25 = 25 x2 + y2 + 12x -10y + 36 = 0 1 3 .) a) Area = 2 1 0510 0250   = 2 1 )2()25(  1 = 2 23 unit2 1 b) C = ( 5 )2(2)5(3  , 5 )5(2)1(3  1 = ( 5 11 , 5 7 ) 1 c) i) Since PA = 2PB 22 )5()2(  yx = 2 22 )1()5(  yx 1 x2 + 4x + 4 + y2 10y + 25 = 4 (x2  10x + 25 + y2 +2y + 1) 1
  10. 10. CHAPTER 6 COORDINATE GEOMETRY FORM 4 29 x2 + y2 + 4x 10y + 29 = 4x2 + 4y2 40x + 8y + 104 3x2 + 3y2 44x + 18y + 75 = 0 1 (ii) When it intersects the y-axis, x = 0.  3y2 +1 8y + 75 = 0 1 Use b2  4ac = (18)2  4(3)(75) 1 = 576 b2  4ac < 0 It does not cut the y-axis since there is no real root. 1 4. a) y + 3x + 9 = 0 When y = 0, 0 + 3x + 9 = 0 x = –3  P(–3, 0) When x = 0, y + 0 + 9 = 0 y = –9  Q(0, –9) 1 R(x, y) = ( 3 )3(2)0(1  , 3 )0(2)9(1  ) 1 = (-2 , -3 ) 1 b) y + 3x + 9 = 0 y = -3x - 9  Gradient of PQ , m1 = –3 1 Since PQ is perpendicular to the straight line, then m1 m2 = 1 Thus, 3 1 2 m The equation of straight line that passes through R(-2, -3) and perpendicular to PQ is 2 3   x y = 3 1 1 3y = x - 7 1 5. a) Equation of the locus of W, 22 )1()5(  yx = 2 5 1 (x – 5)2 + ( y – 1)2 = ( 2 5 )2 1 x2 -10x +25 + y2 – 2y + 1 = 4 25 4 x2 + 4y2 – 40x - 8y + 79 = 0 1 b) i) P(3 , k) lies on the locus of W, substitute x =3 and y = k into the equation of the locus of W. 4(3)2 + 4(k)2 – 40(3) – 8(k) + 79 = 0 1
  11. 11. CHAPTER 6 COORDINATE GEOMETRY FORM 4 30 4k2 - 8k -5 = 0 (2k + 1)(2k – 5) = 0 k = - 2 1 , k = 2 5 Since k > 0,  k = 2 5 1 1 ii) Since S is the centre of the locus of W, then S is the mid-point of PQ. S(5 , 1) = ( 2 3x , 2 2 5 y ) 1 5 = 2 3x , 1 = 2 2 5 y x = 7 , y = - 2 1 Hence, the coordinates of Q are ( 7 , - 2 1 ). 1 c) Area of triangle OPQ = 2 1 0 2 5 2 1 0 0370  = 2 1 [ (7)( 2 5 ) – (- 2 3 ) ] 1 = 2 19 unit2 1

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