Graphs 2

6 marks · Hard

The line L passes through P (–4, 5) and Q (2, –1). A second line M is parallel to L and passes through the point R (1, 7).

(a)

Find the equation of the line L in the form y = mx + c.

[3 marks]

(b)

Write down the gradient of line M.

[1 mark]

(c)

Find the equation of line M in the form y = mx + c.

[2 marks]

Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.

Show answer

(a) y = -x + 1

(b) -1

Also accept: Answers in any correct equivalent form

Common mistakes: Mixing up x and y coordinates when calculating the gradient; Using the wrong point to substitute for c

Show mark scheme and worked solution

Marking points

M1 (a) · 1 mark · Correctly finds the gradient using (–4, 5) and (2, –1).

M2 (a) · 1 mark · Substitutes one point into y = mx + c to find c. (follow-through)

A1 (a) · 1 mark · Correct final equation in the form y = -x + 1. (follow-through)

B1 (b) · 1 mark · Writes down the correct gradient for the parallel line.

M3 (c) · 1 mark · Uses point (1, 7) and correct gradient to attempt equation. (follow-through)

A2 (c) · 1 mark · Correct equation y = -x + 8. (follow-through)

Worked solution

a) Gradient of L:

Substitute (–4, 5) into y = –x + c:

Equation: y = -x + 1

b) Gradient for line M is the same as L since they are parallel: –1

c) Equation of M: use y = –x + c, substitute (1, 7):

Equation: y = -x + 8