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Coordinate Geometry

Coordinate Geometry. 2002 Paper 2 Question 2. 3x - 2y = 12 ---- (2). To find a point of intersection of 2 lines solve by doing a simultaneous equation. 4x + y = 5 ----- (1). Equalise the y terms. 3x - 2y = 12 ----- (2). 8x + 2y = 10 ----. (1) x 2. Add up.

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Coordinate Geometry

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  1. Coordinate Geometry 2002 Paper 2 Question 2

  2. 3x - 2y = 12 ---- (2) To find a point of intersection of 2 lines solve by doing a simultaneous equation 4x + y = 5 ----- (1) Equalise the y terms 3x - 2y = 12 ----- (2) 8x + 2y = 10 ---- (1) x 2 Add up 11x = 22 ----- (1) + (2) x = 2

  3. Substitute 2 for x in (1) Check the results 4x + y = 4(2) + y = 5 5 4(2) + (-3) = 8 + y = 5 8 - 3 = 5 y = 5 - 8 3x – 2y = y = - 3 3(2) – 2(- 3) = 12 x = 2 y = - 3 6 + 6 = 12

  4. Testing a(0, 8) Testing b(-10, 0) 4x – 5y = - 40 4x – 5y = - 40 4(0) – 5(8) = 4(-10) – 5(0) = - 40 - 40 a(0, 8) is on L b(-10, 0) is on L

  5. x1 = 0 y1 = 8 x2 = -10 y2 = 0 y2 – y1 Slope of L = x2 – x1 (0) – (8) = (-10) – (0) -8 = -10 4 = 5

  6. 4 -5 Slope of L = Slope of K = 5 4 -5 x1 = -10 y1 = 0 m = 4 -5 (x – (-10)) y – 0 = 4 4. -5 4y = (x – (-10)) 4 Equation of Line: y – y1 = m(x – x1) K  L Multiply by LCD = 4 b(-10, 0)  K = -5(x + 10) = -5x - 50

  7. 4y = - 5x - 50 5x + 4y + 50 = 0 Equation of K

  8. - 25 = 2 - 50 y = 4 5x + 4y + 50 = 0 --- Equation of K K intersects y-axis when x = 0 5(0) + 4y + 50 = 0 4y = - 50 = - 12½ C is (0, - 12½)

  9. a(0, 8) 8 b(-10, 0) 12.5 c(0, -12.5) Area of abc = ½ b h = ½ (20.5)(10) = 102.5 10 20.5 d Area of abcd = 2(102.5) = 205cm2

  10. a(0, 8) b(-10, 0)  c(0, -12.5) bc maps b(-10, 0)  c(0, -12.5) The translation that maps b onto c also maps a onto d d a(0,8)  d( 10, - 4.5)

  11. Coordinate Geometry 2001 Paper 2 Question 2

  12. 3x + 2y + 7 = 0 (t, 2t) is on the line  3(t) + 2(2t) + 7 = 0  3t + 4t + 7 = 0  7t = - 7  t = - 1

  13. x1 = 4 y1 = 2 x2 = 0 y2 = 4 y2 – y1 Slope of ac = x2 – x1 (4) – (2) = (0) – (4) 2 = - 4 a(4, 2) b(-2, 0) c(0, 4)

  14. x1 = - 2 y1 = 0 x2 = 0 y2 = 4 y2 – y1 Slope of bc = x2 – x1 (4) – (0) = (0) – (-2) 4 = 2 a(4, 2) b(- 2, 0) c(0, 4) = 2

  15. Slope of ac 1 Product of slopes = 2 x - 2 1 = - 2 Slope of bc = 2 = -1  ac  bc

  16. a(4, 2), b(-2, 0) c = (0, 4) x1 = 4 x2 = 0 y1 = 2 y2 = 4

  17. a(4, 2), b(-2, 0) c = (0, 4) x1 = -2 x2 = 0 y1 = 0 y2 = 4  |ac| = |bc|

  18. c(0, 4) a(4, 2) b(-2, 0) b(-2, 0) o Using the translation b(-2, 0)  o(0, 0) a(4, 2)  p(6, 2) c(0, 4)  p(2, 4)

  19. q(2, 4) c(0, 4) a(4, 2) p(6, 2) b(-2, 0) o Area of abc = Area of opq = ½|x1y2 – x2y1| x1 = 6 x2 = 2 y1 = 2 y2 = 4

  20. x1 = 6 x2 = 2 y1 = 2 y2 = 4 Area of abc = Area of opq = ½|x1y2 – x2y1| = ½|(6)(4) – (2)(2)| = ½|24 – 4| = ½(20) = 10 cm2

  21. g h c(0, 4) b(-2, 0) a(4, 2) The translation that maps b(-2, 0)  c(0, 4) will also map c(0, 4)  h (2, 8)

  22. g h c(0, 4) b(-2, 0) a(4, 2) The translation that maps a(4, 2)  c(0, 4) will also map c(0, 4)  g (-4, 6)

  23. x1 = -2 y1 = 0 x2 = 0 y2 = 4 y2 – y1 Slope of bc = x2 – x1 (4) – (0) = (0) – (-2) 4 = 2 b(-2, 0) c(0, 4) = 2

  24. b(-2, 0) c(0, 4) Slope = 2 x1 = 0 m = 2 y1 = 4 Equation of bc: y – y1 = m(x – x1) y – 4 = 2(x – 0) y – 4 = 2x 2x – y + 4 = 0 Equation of bc

  25. Coordinate Geometry 2000 Paper 2 Question 2

  26. 2x – y + 4 = 0 Equation of bc h(2, 8) When x = 2 and y = 8 2(2) – (8) + 4 = 0  h(2, 8) is on the line bc

  27. a(2, -3) b(-8, -6) x1 = 2 x2 = -8 y1 = -3 y2 = -6 Mid-point of [ab] =

  28. a(-2, -1), b(1, 0) c = (-5, 2) x1 = -2 x2 = 1 y1 = -1 y2 = 0

  29. a(-2, -1), b(1, 0) c = (-5, 2) x1 = 1 x2 = -5 y1 = 0 y2 = 2

  30. |ab| : |bc| = = 1 : 2

  31. L: 3x – 4y + 20 = 0 To find the slope of L get y on it’s own - 4y = - 3x - 20 Change Signs Divide by 4 4y = 3x + 20 Slope of L = No. in front of x y = ¾x + 5 Slope of L = ¾

  32. Slope of K x1 = 0 y1 = 5 y – (5) = (x – 0) 3y – 15 = x Slope of L = ¾ K  L Equation of K: y – y1 = m(x – x1) 3y – 15 = - 4x Multiply by LCD = 3 4x + 3y – 15 = 0 K: 4x + 3y – 15 = 0

  33. x = 15 x = - 20 4 3 15 - 20 , 0) t = ( r = ( , 0) 4 3 K: 4x + 3y – 15 = 0 L: 3x – 4y + 20 = 0 L cuts x-axis when y = 0 K cuts x-axis when y = 0 4x + 3(0) – 15 = 0 3x – 4(0) + 20 = 0 4x = 15 3x = - 20

  34. p(0, 5) 15 , 0) r ( 4 x 5 = ½ x 625 125 125 = - 20 t ( , 0) 24 12 12 3 5 4 3 5 2 1 0 Area of ptr = ½ b.h

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