1. Coordinate Geometry 2002 Paper 2 Question 2

3. To find a point of intersection of 2 lines solve by doing a simultaneous equation 4x + y = 5 ----- (1) 3x - 2y = 12 ----- (2) Equalise the y terms (1) x 28x + 2y = 10 ---- 3x - 2y = 12 ---- (2) Add up 11x = 22 ----- (1) + (2) x = 2

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

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

6. Slope of L = y 2 – y 1 x 2 – x 1 x 1 = 0 y 1 = 8 x 2 = -10 y 2 = 0 (0) – (8) (-10) – (0) = -8-8 -10 = 4 5 =

7. Slope of L = 4 5 K  L Slope of K = -5 4 b(-10, 0)  K Equation of Line: y – y 1 = m(x – x 1 ) m = -5 4 x 1 = -10 y 1 = 0 y – 0 = -5 4 (x – (-10)) Multiply by LCD = 4 4y = -5 4 (x – (-10)) 4. = -5(x + 10) = -5x - 50

8. 5x + 4y + 50 = 0 Equation of K 4y = - 5x - 50

9. 5x + 4y + 50 = 0 --- Equation of K K intersects y-axis when x = 0 5(0) + 4y + 50 = 0 4y = - 50 y = - 50 4 - 25 2 = = - 12½ C is (0, - 12½)

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

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

12. Coordinate Geometry 2001 Paper 2 Question 2

14. 3x + 2y + 7 = 0 (t, 2t) is on the line  3(t) + 2(2t) + 7 = 0  3t + 4t + 7 = 0  7t = - 7  t = - 1

15. Slope of ac = y 2 – y 1 x 2 – x 1 x 1 = 4 y 1 = 2 x 2 = 0 y 2 = 4 (4) – (2) (0) – (4) = 2 - 4 = a(4, 2) b(-2, 0) c(0, 4)

16. Slope of bc = y 2 – y 1 x 2 – x 1 x 1 = - 2 y 1 = 0 x 2 = 0 y 2 = 4 (4) – (0) (0) – (-2) = 4 2 = a(4, 2) b(- 2, 0) c(0, 4) = 2

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

18. a(4, 2), b(-2, 0) c = (0, 4) x 1 = 4 x 2 = 0 y 1 = 2 y 2 = 4

19. a(4, 2), b(-2, 0) c = (0, 4) x 1 = -2 x 2 = 0 y 1 = 0 y 2 = 4  |ac| = |bc|

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

21. a(4, 2) c(0, 4) o b(-2, 0) p(6, 2) q(2, 4) Area of  abc = Area of  opq = ½|x 1 y 2 – x 2 y 1 | x 1 = 6 x 2 = 2 y 1 = 2 y 2 = 4

22. Area of  abc = Area of  opq = ½|x 1 y 2 – x 2 y 1 | x 1 = 6 x 2 = 2 y 1 = 2 y 2 = 4 = ½|(6)(4) – (2)(2)| = ½|24 – 4| = ½(20) = 10 cm 2

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

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

25. Slope of bc = y 2 – y 1 x 2 – x 1 x 1 = -2 y 1 = 0 x 2 = 0 y 2 = 4 (4) – (0) (0) – (-2) = 4 2 = b(-2, 0) c(0, 4) = 2

26. b(-2, 0) c(0, 4) Slope = 2 Equation of bc: y – y 1 = m(x – x 1 ) x 1 = 0 m = 2 y 1 = 4 y – 4 = 2(x – 0) y – 4 = 2x 2x – y + 4 = 0 Equation of bc

27. Coordinate Geometry 2000 Paper 2 Question 2

28. 2x – y + 4 = 0 Equation of bch(2, 8) When x = 2 and y = 8 2(2) – (8) + 4 =0  h(2, 8) is on the line bc

29. a(2, -3) b(-8, -6) Mid-point of [ab] = x 1 = 2 x 2 = -8 y 1 = -3 y 2 = -6

30. a(-2, -1), b(1, 0) c = (-5, 2) x 1 = -2 x 2 = 1 y 1 = -1 y 2 = 0

31. a(-2, -1), b(1, 0) c = (-5, 2) x 1 = 1 x 2 = -5 y 1 = 0 y 2 = 2

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

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

34. K  L Slope of K Slope of L = ¾ Equation of K: y – y 1 = m(x – x 1 ) x 1 = 0 y 1 = 5 y – (5) = (x – 0) Multiply by LCD = 3 3y – 15 = x 3y – 15 = - 4x 4x + 3y – 15 = 0 K: 4x + 3y – 15 = 0

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

36. 1 2 3 4 5 0 p(0, 5) Area of  ptr = ½ b.h 125 12 5 = ½ x 125 12 x 5 625 24 = - 20 3 t (, 0) 15 4 r (, 0)