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AlgebraFoundation and Higher (some Higher only)

Straight-line graphs GCSE questions with answers

8 straight-line graphs questions to try, each with the answer and the full working one tap away. Every question here was made and worked out by the Show Working maths engine, not typed in by hand.

Straight-line graph questions use y = mx + c, where m is the gradient and c is where the line crosses the y-axis. You find gradients, intercepts, equations of lines and midpoints, and on Higher you also work with parallel and perpendicular lines.

Quiz yourself: 8 questions

Worked example 1: gradient and y-intercept

A is the point (−3, −3) and B is the point (−1, −5). Work out the gradient of the line AB.

  1. Change in y: −5 − (−3) = −2.
  2. Change in x: −1 − (−3) = 2.
  3. Gradient = change in y ÷ change in x = −22 = −1.

Answer −1

01 / Quiz yourselfTry it, then tap to check

Straight-line graphs questions to try

Work each one out on paper first. Then open the answer and compare your working with ours, step by step.

01 / Gradient and y-interceptBoth tiers

A is the point (−3, 6) and B is the point (−7, 10). Work out the gradient of the line AB.

Show the answer and working
  1. Change in y: 10 − 6 = 4.
  2. Change in x: −7 − (−3) = −4.
  3. Gradient = change in y ÷ change in x = 4−4 = −1.

Answer −1

02 / Gradient and y-interceptBoth tiers

A is the point (−2, −8) and B is the point (−5, −23). Work out the gradient of the line AB.

Show the answer and working
  1. Change in y: −23 − (−8) = −15.
  2. Change in x: −5 − (−2) = −3.
  3. Gradient = change in y ÷ change in x = −15−3 = 5.

Answer 5

03 / Midpoint of a line segmentBoth tiers

A is the point (−8, 4) and B is the point (−7, −3). Find the coordinates of the midpoint of AB.

Show the answer and working
  1. The midpoint is halfway between the two points: add each pair of coordinates and halve.
  2. x: −8 + (−7)2 = −7.5.
  3. y: 4 + (−3)2 = 0.5.
  4. The midpoint is (−7.5, 0.5).

Answer (−7.5, 0.5)

04 / Midpoint of a line segmentBoth tiers

M is the midpoint of the line AB. A is the point (−5, −1) and M is the point (−0.5, 0). Find the coordinates of B.

Show the answer and working
  1. To get from A to M: x changes by −0.5 − (−5) = 4.5 and y changes by 0 − (−1) = 1.
  2. B is the same step again from M: x = −0.5 + 4.5 = 4, y = 0 + 1 = 1.
  3. So B is (4, 1).

Answer (4, 1)

05 / Equation of a straight lineBoth tiers

A straight line has gradient 1 and passes through the point (2, 0).
Find the equation of the line. Give your answer in the form y = mx + c.

Show the answer and working
  1. Put the point (2, 0) into y = x + c: 0 = 1 × 2 + c.
  2. So c = 0 − 2 = −2.
  3. The equation is y = x − 2.

Answer y = x − 2

06 / Equation of a straight lineBoth tiers

A straight line passes through the points (6, −8) and (−3, −2).
Find the equation of the line. Give your answer in the form y = mx + c.

Show the answer and working
  1. Gradient = change in y ÷ change in x = −2 − (−8)−3 − 6 = −23.
  2. Put the point (6, −8) into y = −23x + c: −8 = −23 × 6 + c.
  3. So c = −8 − (−4) = −4.
  4. The equation is y = −23x − 4.

Answer y = −23x − 4

07 / Parallel and perpendicular linesHigher only

Line L is parallel to the line y = 3x + 5 and passes through the point (3, 1).
Find the equation of line L. Give your answer in the form y = mx + c.

Show the answer and working
  1. Rearrange y = 3x + 5 into the form y = mx + c: y = 3x + 5. Its gradient is 3.
  2. Parallel lines have the same gradient, so L has gradient 3.
  3. Put (3, 1) into y = 3x + c: 1 = 3 × 3 + c, so c = −8.
  4. So line L is y = 3x − 8.

Answer y = 3x − 8

08 / Parallel and perpendicular linesHigher only

Line L is perpendicular to the line 2y − x = −4 and passes through the point (−4, −5).
Find the equation of line L. Give your answer in the form y = mx + c.

Show the answer and working
  1. Rearrange 2y − x = −4 into the form y = mx + c: y = 12x − 2. Its gradient is 12.
  2. Perpendicular gradients multiply to −1, so the new gradient is −1 ÷ 12 = −2. (Flip the fraction and change the sign.)
  3. Put (−4, −5) into y = −2x + c: −5 = −2 × (−4) + c, so c = −13.
  4. So line L is y = −2x − 13.

Answer y = −2x − 13

02 / What's covered4 kinds of question

Straight-line graphs at GCSE and where marks go

  • Gradient and y-interceptBoth tiers
  • Midpoint of a line segmentBoth tiers
  • Equation of a straight lineBoth tiers
  • Parallel and perpendicular linesHigher only

Worked example 2: parallel and perpendicular lines

Line L is parallel to the line 2y + 6x = −8 and passes through the point (−4, −1).
Find the equation of line L. Give your answer in the form y = mx + c.

  1. Rearrange 2y + 6x = −8 into the form y = mx + c: y = −3x − 4. Its gradient is −3.
  2. Parallel lines have the same gradient, so L has gradient −3.
  3. Put (−4, −1) into y = −3x + c: −1 = −3 × (−4) + c, so c = −13.
  4. So line L is y = −3x − 13.

Answer y = −3x − 13

A common mistake

Reading the gradient before y is on its own. For 2y = 6x + 4, divide everything by 2 first to get y = 3x + 2, so the gradient is 3, not 6.

03 / Practise itiPhone and iPad

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