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Practise 19 exam-style questions on Translations as vectors 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.
Shape A is mapped onto shape B by the translation \(\begin{pmatrix}2\\-5\end{pmatrix}\).
Write down the column vector of the translation that maps shape B onto shape A.
\(\begin{pmatrix}-2\\5\end{pmatrix}\)
The inverse of a translation moves the same distance in the opposite direction, so both components change sign.
B1 for \(\begin{pmatrix}-2\\5\end{pmatrix}\)
accept (−2, 5) typed as a vector
The grid shows triangle A and triangle B.
Kai says, "The translation that maps triangle A onto triangle B is \(\begin{pmatrix}-3\\2\end{pmatrix}\)."
Explain the mistake Kai has made and write down the correct vector.
Kai's vector maps B onto A, not A onto B: from A to B every vertex moves 3 right and 2 down, so both of his signs are wrong.
The correct vector is \(\begin{pmatrix}3\\-2\end{pmatrix}\).
Always measure the movement from the object (A) to the image (B). Moving right is positive in the top number and moving down is negative in the bottom number.
C1 for explaining that Kai has given the translation from B to A (the signs are the wrong way round / he has gone in the opposite direction)
oe
B1 for \(\begin{pmatrix}3\\-2\end{pmatrix}\)
accept (3, −2) typed as a vector
The translation \(\begin{pmatrix}a + 1\\2b\end{pmatrix}\) maps the point (3, 1) onto the point (−1, 7).
Work out the value of \(a\) and the value of \(b\).
The vector from (3, 1) to (−1, 7) is \(\begin{pmatrix}-1 - 3\\7 - 1\end{pmatrix} = \begin{pmatrix}-4\\6\end{pmatrix}\)
\(a + 1 = -4\), so \(a = -5\)
\(2b = 6\), so \(b = 3\)
Work out the actual translation vector from the two points, then set each component equal to the algebraic component and solve the two separate equations.
M1 for the translation vector \(\begin{pmatrix}-4\\6\end{pmatrix}\)
or \(3 + a + 1 = -1\) or \(1 + 2b = 7\); correct answer with no working scores full marks
A1 for \(a = -5\)
A1 for \(b = 3\)
The grid shows triangle ABC.
Triangle ABC is translated by the vector \(\begin{pmatrix}p\\q\end{pmatrix}\) to give triangle A′B′C′.
C′ lies on the \(x\)-axis and A′ lies on the line \(y = x\).
Work out the value of \(p\) and the value of \(q\).
From the grid: A(1, 2), B(4, 2), C(1, 6)
C′ = \((1 + p, 6 + q)\) lies on the \(x\)-axis, so \(6 + q = 0\) and \(q = -6\)
A′ = \((1 + p, 2 + q) = (1 + p, -4)\) lies on \(y = x\), so \(1 + p = -4\) and \(p = -5\)
\(p = -5\), \(q = -6\)
Write the image of each vertex in terms of \(p\) and \(q\). A point on the \(x\)-axis has \(y\)-coordinate 0, and a point on \(y = x\) has equal coordinates. Check: A′ = (−4, −4), B′ = (−1, −4) and C′ = (−4, 0).
P1 for C′ = \((1 + p, 6 + q)\) or A′ = \((1 + p, 2 + q)\)
using A(1, 2) and C(1, 6) read from the grid; correct answer with no working scores full marks
P1 for \(6 + q = 0\), so \(q = -6\)
P1 for \(1 + p = 2 + q\)
A′ on \(y = x\); ft their \(q\)
A1 for \(p = -5\) and \(q = -6\)
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