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Practise 28 exam-style questions on Straight-line graphs and y = mx + c for Edexcel GCSE Maths (Non-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.
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.
Write down the equation of the straight line that passes through the point \((4, -3)\) and is parallel to the y-axis.
\(x = 4\)
A line parallel to the y-axis is vertical, so every point on it has the same x-coordinate. Here that is 4, so the equation is \(x = 4\). A common mistake is to give \(y = -3\), which is the horizontal line through the point.
B1 for \(x = 4\)
cao; do not accept \(y = -3\)
The line with equation \(y = kx + 3\) is perpendicular to the line with equation \(2x + 5y = 1\).
Find the value of \(k\).
\(5y = -2x + 1\)
\(y = -\dfrac{2}{5}x + \dfrac{1}{5}\), so the gradient is \(-\dfrac{2}{5}\).
Perpendicular gradient \(= \dfrac{5}{2}\)
\(k = \dfrac{5}{2}\)
Make \(y\) the subject to find the gradient of the given line, then take the negative reciprocal. A common mistake is to read the gradient as 2 or \(-2\) straight from the equation without rearranging.
M1 for finding the gradient of \(2x + 5y = 1\) as \(-\frac{2}{5}\)
e.g. from \(y = -\frac{2}{5}x + \frac{1}{5}\); correct answer with no working scores full marks
A1 for \(k = \frac{5}{2}\)
oe, e.g. 2.5
\(A\) is the point \((1, 3)\), \(B\) is the point \((3, 7)\) and \(C\) is the point \((6, 13)\).
Prove that \(A\), \(B\) and \(C\) lie on a straight line.
Gradient of \(AB = \dfrac{7 - 3}{3 - 1} = \dfrac{4}{2} = 2\)
Gradient of \(BC = \dfrac{13 - 7}{6 - 3} = \dfrac{6}{3} = 2\)
\(AB\) and \(BC\) have the same gradient and both pass through \(B\), so \(A\), \(B\) and \(C\) lie on the same straight line.
Equal gradients alone only show that two lines are parallel; you also need a common point, here \(B\), to show they are the same line. Another method is to find the equation of \(AB\), \(y = 2x + 1\), and show that \(C\) fits it: \(2 \times 6 + 1 = 13\).
M1 for finding the gradient of one of \(AB\), \(BC\) or \(AC\), e.g. \(\dfrac{7 - 3}{3 - 1} = 2\)
or for the equation of AB, \(y = 2x + 1\)
M1 for finding a second gradient, e.g. \(\dfrac{13 - 7}{6 - 3} = 2\), or substituting \(C\) into the equation of \(AB\)
C1 for a complete argument: both gradients are 2 and the lines share the point \(B\) (or \(C\) satisfies \(y = 2x + 1\)), so the points are on a straight line
The line \(L_1\) has equation \(y = 2x + 1\).
\(L_2\) is perpendicular to \(L_1\) and passes through the point \((8, 2)\).
\(L_1\) and \(L_2\) intersect at the point \(P\).
Find the coordinates of \(P\).
Gradient of \(L_2 = -\dfrac{1}{2}\)
\(2 = -\dfrac{1}{2} \times 8 + c\), so \(c = 6\)
\(L_2\): \(y = -\dfrac{1}{2}x + 6\)
At \(P\): \(2x + 1 = -\dfrac{1}{2}x + 6\)
\(\dfrac{5}{2}x = 5\)
\(x = 2\), \(y = 2 \times 2 + 1 = 5\)
\(P = (2, 5)\)
First find the equation of \(L_2\) using the negative reciprocal gradient and the point \((8, 2)\). The intersection is where both equations give the same \(y\), so set them equal and solve. Check \(P\) in \(L_2\): \(-\frac{1}{2} \times 2 + 6 = 5\).
M1 for the gradient of \(L_2\) as \(-\frac{1}{2}\)
correct answer with no working scores full marks
M1 for the equation of \(L_2\), \(y = -\frac{1}{2}x + 6\)
ft their gradient; oe
M1 for setting \(2x + 1 = -\frac{1}{2}x + 6\) and solving to find \(x\)
ft their equation of \(L_2\)
A1 for \(P = (2, 5)\)
cao
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