Parallel and perpendicular lines

Mark scheme and answers · Total marks: 101

Rate this booklet

How would you rate this worksheet overall (questions, layout, and answers together)? Tap the stars from the left up to the score you mean, then press Confirm rating. You can change your mind any time.

Average: 4.5 / 5 · sample

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).