Straight-line graphs and y = mx + c - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 28 exam-style questions on Straight-line graphs and y = mx + c for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

28 questions
1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

The table shows some values of \(x\) and \(y\) for the straight line with equation \(y = 3.5 - 1.5x\)

Find the value of \(a\) and the value of \(b\).

Table with columns "x", "−2", "−1", "0", "1", "2", "3". x: y; −2: 6.5; −1: a; 0: 3.5; 1: 2; 2: b; 3: −1.

Model answer

\(x = -1\): \(y = 3.5 - 1.5 \times (-1) = 3.5 + 1.5 = 5\), so \(a = 5\)
\(x = 2\): \(y = 3.5 - 1.5 \times 2 = 3.5 - 3 = 0.5\), so \(b = 0.5\)

Explanation

Substitute each x-value into the equation. Take care with \(-1.5 \times (-1) = +1.5\). You can check with the pattern in the table: each time \(x\) goes up by 1, \(y\) goes down by 1.5, because the gradient is \(-1.5\).

Mark scheme

1 mark
B1 for \(a = 5\)
cao; correct answer with no working scores full marks
1 mark
B1 for \(b = 0.5\)
cao

Question 2

3 marks

A straight line passes through the points \((-3.5, 6.2)\) and \((1.5, -1.8)\).

Find the equation of the line.
Give your answer in the form \(y = mx + c\)

Model answer

Gradient \(= \dfrac{-1.8 - 6.2}{1.5 - (-3.5)} = \dfrac{-8}{5} = -1.6\)
\(y = -1.6x + c\)
At \((1.5, -1.8)\): \(-1.8 = -1.6 \times 1.5 + c = -2.4 + c\)
\(c = 0.6\)
\(y = -1.6x + 0.6\)

Explanation

Find the gradient from the change in y divided by the change in x, subtracting the coordinates in the same order. Then substitute either point to find \(c\), and check with the other point: \(-1.6 \times (-3.5) + 0.6 = 6.2\).

Mark scheme

1 mark
M1 for a correct method to find the gradient, e.g. \(\dfrac{-1.8 - 6.2}{1.5 - (-3.5)}\)
oe; correct answer with no working scores full marks
1 mark
M1 for substituting a point into \(y = mx + c\) with their gradient, e.g. \(-1.8 = -1.6 \times 1.5 + c\)
oe, e.g. \(y - 6.2 = -1.6(x + 3.5)\)
1 mark
A1 for \(y = -1.6x + 0.6\)
oe (exact)

Question 3

3 marks

The diagram shows the lines \(L_1\) and \(L_2\) drawn on a grid.

Show that \(L_1\) and \(L_2\) are perpendicular.

A coordinate grid with x from −4 to 4 and y from −4 to 4; each grid square is 1 unit across, and the axes cross at the origin O. Line L₁ is a straight line sloping upwards from left to right passing through (−3, −2) and (3, 2). Line L₂ is a straight line sloping downwards from left to right passing through (−2, 3) and (2, −3). Line L₁ and line L₂ cross at (0, 0).

Model answer

\(L_1\) passes through \((-3, -2)\) and \((3, 2)\): gradient \(= \dfrac{2 - (-2)}{3 - (-3)} = \dfrac{4}{6} = \dfrac{2}{3}\)
\(L_2\) passes through \((-2, 3)\) and \((2, -3)\): gradient \(= \dfrac{-3 - 3}{2 - (-2)} = \dfrac{-6}{4} = -\dfrac{3}{2}\)
\(\dfrac{2}{3} \times \left(-\dfrac{3}{2}\right) = -1\)
The product of the gradients is \(-1\), so \(L_1\) and \(L_2\) are perpendicular.

Explanation

Two lines are perpendicular when the product of their gradients is \(-1\). Read the gradients from points where each line crosses grid intersections exactly. Lines that merely look perpendicular on a diagram are not a proof.

Mark scheme

1 mark
M1 for a correct method to find the gradient of \(L_1\) or \(L_2\), e.g. \(\dfrac{4}{6}\) or \(\dfrac{-6}{4}\)
1 mark
A1 for both gradients: \(L_1\): \(\frac{2}{3}\) and \(L_2\): \(-\frac{3}{2}\)
oe
1 mark
C1 for \(\frac{2}{3} \times \left(-\frac{3}{2}\right) = -1\), so the lines are perpendicular
or \(-\frac{3}{2}\) is the negative reciprocal of \(\frac{2}{3}\); answer only scores 0

Question 4

4 marks

The diagram shows the straight line \(L\) drawn on a grid.

The line \(M\) is parallel to \(L\) and passes through the point \((5, -1)\).

Find the coordinates of the point where \(M\) crosses the x-axis.

A coordinate grid with x from −5 to 5 and y from −2 to 5; each grid square is 1 unit across, and the axes cross at the origin O. Line L is a straight line sloping upwards from left to right passing through (−2, 1) and (4, 4).

Model answer

\(L\) passes through \((-4, 0)\) and \((2, 3)\): gradient \(= \dfrac{3}{6} = 0.5\)
\(M\) is parallel, so its gradient is 0.5
\(-1 = 0.5 \times 5 + c\), so \(c = -3.5\)
\(M\): \(y = 0.5x - 3.5\)
When \(y = 0\): \(0.5x = 3.5\), \(x = 7\)
\(M\) crosses the x-axis at \((7, 0)\)

Explanation

Parallel lines have the same gradient, so read the gradient of \(L\) from the grid and use it with the point \((5, -1)\). Then set \(y = 0\) for the x-axis crossing. Quick check: from \((5, -1)\), going 2 right and 1 up reaches \((7, 0)\).

Mark scheme

1 mark
M1 for the gradient of \(L\), \(\frac{1}{2}\)
oe, e.g. \(\frac{3}{6}\); correct answer with no working scores full marks
1 mark
M1 for substituting \((5, -1)\) with their gradient, e.g. \(-1 = 0.5 \times 5 + c\)
oe, e.g. \(y + 1 = 0.5(x - 5)\)
1 mark
M1 for \(y = 0.5x - 3.5\) and substituting \(y = 0\)
oe
1 mark
A1 for (7, 0)
cao

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