Parallel and perpendicular lines

Mark scheme and answers · Total marks: 101

Question 1 [5 marks]

(a) Any line of the form y = 12x + k with k ≠ 2 (e.g. y = 12x + 1).
(b) Any line through (0, 2) except AB itself (e.g. y = 3x + 2 or y = 2).
(c) Gradient of AB is 12, so perpendicular gradient is -2. Through B(8, 6): y - 6 = -2(x - 8), hence y = -2x + 22.

Question 2 [2 marks]

(a) R (gradient 1 matches y = x + 6).
(b) S (gradient -12 is negative reciprocal of 2).

Question 3 [1 mark]

5

Question 4 [1 mark]

-3

Question 5 [2 marks]

y = -4x (or y = -4x + 0)

Question 6 [3 marks]

Gradient of given line is 2, so perpendicular gradient is -12. Through (0, 5): y = -12x + 5.

Question 7 [3 marks]

Perpendicular gradient is -13. Through (3, 8): y - 8 = -13(x - 3), hence y = -13x + 9.

Question 8 [3 marks]

First line: y = 12x + 2, gradient 12. Second gradient -2. Product 12 × (-2) = -1, so the lines are perpendicular.

Question 9 [2 marks]

Gradient PQ = (11 - 3)(5 - 1) = 2. A parallel line has the same gradient: 2.

Question 10 [2 marks]

y = -3 (horizontal). A perpendicular line is vertical: x = 4.

Question 11 [4 marks]

3y = 4x - 12 gives y = 43x - 4, gradient 43. 4y = -3x + 20 gives y = -34x + 5, gradient -34. Product 43 × (-34) = -1, hence perpendicular.

Question 12 [4 marks]

(a) Parallel to CD has gradient 1 through (0, 0): y = x.
(b) Perpendicular gradient -1 through D(6, 9): y - 9 = -(x - 6), hence y = -x + 15.

Question 13 [2 marks]

(a) P or R (both gradient 3).
(b) Q (gradient -13 is negative reciprocal of 13).

Question 14 [3 marks]

Perpendicular gradient is -12. Through (4, 7): y - 7 = -12(x - 4), hence y = -12x + 9.

Question 15 [3 marks]

2x + 4y = 12 gives y = -12x + 3, gradient -12. Parallel through (0, 1): y = -12x + 1 (or x + 2y = 2).

Question 16 [3 marks]

Gradient EF = (10 - 2)(3 - (-1)) = 2. Gradient GH = k3. Parallel gives k3 = 2, so k = 6.

Question 17 [4 marks]

Midpoint (5, 8). Gradient of segment = (10 - 6)(8 - 2) = 23. Perpendicular gradient = -32. Equation: y - 8 = -32(x - 5), hence y = -32x + 312.

Question 18 [4 marks]

At (2, 5): 5 = 4 + c \Rightarrow c = 1, and 5 = -1 + d \Rightarrow d = 6. Gradients 2 and -12 multiply to -1, so perpendicular.

Question 19 [3 marks]

Gradients 34 and -43. Product 34 × (-43) = -1, hence perpendicular.

Question 20 [5 marks]

(a) Gradient 23 through B(7, 5): y - 5 = 23(x - 7), hence y = 23x - 43.
(b) Perpendicular gradient -32 through A(1, 1): y - 1 = -32(x - 1), hence y = -32x + 52.

Question 21 [5 marks]

Radius gradient from (2, 5) to (11, 8) is (8 - 5)(11 - 2) = 13. Tangent is perpendicular, gradient -3. Through (11, 8): y - 8 = -3(x - 11), hence y = -3x + 41.

Question 22 [5 marks]

Radius gradient (6 - 2)(5 - (-1)) = 46 = 23. Tangent gradient -32. Through (5, 6): y - 6 = -32(x - 5), hence y = -32x + 272.

Question 23 [4 marks]

2x - 5y = 10 gives y = 25x - 2, gradient 25. 5x + 2y = 4 gives y = -52x + 2, gradient -52. Product 25 × (-52) = -1.

Question 24 [4 marks]

Write y = -a2x + 4 and y = -3bx + 12b. Parallel gradients: -a2 = -3b, so ab = 6. E.g. a = 2, b = 3 (or a = 1, b = 6, etc.).

Question 25 [3 marks]

3y = 12x - 9 gives y = 4x - 3, gradient 4. Through (4, -1): y + 1 = 4(x - 4), hence y = 4x - 17.

Question 26 [4 marks]

Diagonal gradient (0 - 4)(6 - 0) = -23. Perpendicular gradient 32. Through (0, 0): y = 32x.

Question 27 [3 marks]

Gradients m and -1m. Product m × (-1m) = -1 for all m ≠ 0, so L1 and L2 are always perpendicular.

Question 28 [4 marks]

RS and ST have equal length; symmetry line is perpendicular bisector of RT. R and T have same y; midpoint of RT is (5, 2). RT is horizontal, so symmetry line is vertical x = 5.

Question 29 [5 marks]

Intersection: 2x + 1 = 7 - x gives x = 2, y = 5. Parallel to y = 12x has gradient 12. Through (2, 5): y - 5 = 12(x - 2), hence y = 12x + 4.

Question 30 [5 marks]

(a) Perpendicular gradient -2 through Q(10, 6): y - 6 = -2(x - 10), hence y = -2x + 26.
(b) Through P parallel to y = -2x + 3 is y = -2x + 1. Meets x-axis when y = 0: 0 = -2x + 1, x = 12. R is (12, 0).