Worksheets · Foundation and Higher

Reflections, rotations and translations

8 exam-style questions, grades 2 to 6. Worked solutions and the marks are on the last page.

Question 1Non-calculator · 3 marks
This question is about translations.
(a) The point $A(2, 5)$ is translated to the point $A'(6, 1)$ . Write down the column vector of the translation. (1)
(b) The point $B(-3, 2)$ is translated by the vector $\begin{pmatrix} -2 \\ 5 \end{pmatrix}$ . Write down the coordinates of the image of $B$ . (1)
(c) Write down the column vector that translates $A'$ back to $A$ . (1)
Question 2Non-calculator · 3 marks
(a) Point P has coordinates ( $-3$ , 5). It is translated by the column vector (7, $-9$ ), meaning 7 units right and 9 units down. Write the coordinates of its image. (3)
Question 3Non-calculator · 2 marks
(a) Point Q has coordinates ( $-6$ , 2). Reflect Q in the y-axis. Write the coordinates of its image. (2)
Question 4Non-calculator · 3 marks
Triangle $P$ has vertices $(1, 1)$ , $(3, 1)$ and $(3, 2)$ .
(a) Triangle $P$ is rotated $90^\circ$ clockwise about the origin. Write down the coordinates of the image of the vertex $(3, 2)$ . (1)
(b) Triangle $R$ has vertices $(-1, -1)$ , $(-3, -1)$ and $(-3, -2)$ . Describe fully the single transformation that maps triangle $P$ onto triangle $R$ . (2)
Question 5Non-calculator · 4 marks
Point A is $(3,-2)$ . Rotate A 90 degrees anticlockwise about the origin.
(a) Find the image coordinates. (2)
(b) Find the image after another 90 degree anticlockwise turn. (2)
Question 6Non-calculator · 4 marks
A translation sends A $(4,-3)$ to B $(-1,0)$ .
(a) Find the translation vector. (2)
(b) Find the image of C $(-4,4)$ . (2)
Question 7Non-calculator · 4 marks
Triangle T has vertices (1, 2), (4, 2) and (1, 4).
(a) T is reflected in the line y = x. Write down the image of the vertex (4, 2). (1)
(b) T is rotated 180° about the origin. Write down the image of the vertex (1, 4). (1)
(c) T is translated so that (1, 2) moves to (6, −1). Write the translation vector as (x, y). (2)
Question 8Non-calculator · 5 marks
This question is about reflections.
(a) The point $P(0, 6)$ is reflected in the line $L$ . Its image is $P'(4, 2)$ . Find an equation of the line $L$ . (3)
(b) Triangle $T$ has vertices $(2, 1)$ , $(5, 1)$ and $(5, 3)$ . Triangle $U$ has vertices $(-1, -2)$ , $(-1, -5)$ and $(-3, -5)$ . Describe fully the single transformation that maps $T$ onto $U$ . (2)

Worked solutions and marks

Question 1

(a) $\begin{pmatrix} 4 \\ -4 \end{pmatrix}$

Across: $6 - 2 = 4$ . Up: $1 - 5 = -4$ (down 4).

B1 The vector $(4, -4)$ written as a column.

(b) $(-5, 7)$

$(-3 + (-2),\ 2 + 5) = (-5, 7)$ .

B1 The image $(-5, 7)$ .

(c) $\begin{pmatrix} -4 \\ 4 \end{pmatrix}$

Going back reverses the movement, so both components change sign.

B1 The vector $(-4, 4)$ .

Question 2

(a) $(4,-4)$

$-3+7$
$5-9$
Add the horizontal component: $-3$ + 7 = 4.
Add the vertical component: 5 $-$ 9 = $-4$ , giving (4, $-4$ ).

P1 Establishing $-3+7$ or an equivalent valid method.
P1 Establishing $5-9$ or an equivalent valid method.
A1 Correct answer: $(4,-4)$

Question 3

(a) $(6,2)$

$0-(-6)$
Reflection in the y-axis changes the sign of x and leaves y unchanged.
The image is (6, 2).

P1 Establishing $0-(-6)$ or an equivalent valid method.
A1 Correct answer: $(6,2)$

Question 4

(a) $(2, -3)$

A $90^\circ$ clockwise turn about $O$ sends $(x, y)$ to $(y, -x)$ .
So $(3, 2) \to (2, -3)$ . Check: the point moves from the first quadrant to the fourth.

B1 The image $(2, -3)$ .

(b) Rotation of $180^\circ$ about the origin $(0, 0)$

Each vertex $(x, y)$ goes to $(-x, -y)$ .
That is a half turn: rotation $180^\circ$ about $(0, 0)$ . A half turn needs no direction.

B1 Naming a rotation.
B1 Both $180^\circ$ (a half turn) and the centre $(0, 0)$ .

Question 5

(a) $(2,3)$

A quarter turn anticlockwise sends (x,y) to (-y,x).
$x=-(-2)$
Therefore $(2,3)$ .

M1 A quarter turn anticlockwise sends (x,y) to (-y,x).
A1 Correct answer: $(2,3)$

(b) $(-3,2)$

Apply the same rotation rule a second time.
$x=-3$
Therefore $(-3,2)$ .

M1 Apply the same rotation rule a second time.
A1 Correct answer: $(-3,2)$

Question 6

(a) $(-5,3)$

Subtract the starting coordinates from the final coordinates.
$(-1)-(4)$
Therefore $(-5,3)$ .

M1 Subtract the starting coordinates from the final coordinates.
A1 Correct answer: $(-5,3)$

(b) $(-9,7)$

Apply the same displacement to C.
$-4-5$
Therefore $(-9,7)$ .

M1 Apply the same displacement to C.
A1 Correct answer: $(-9,7)$

Question 7

(a) $(2,4)$

Therefore $(2,4)$ .

B1 Correct answer: $(2,4)$

(b) $(-1,-4)$

Therefore $(-1,-4)$ .

B1 Correct answer: $(-1,-4)$

(c) $(5,-3)$

Subtract the original coordinates from the image coordinates.
$6-1$
Therefore $(5,-3)$ .

M1 Subtract the original coordinates from the image coordinates.
A1 Correct answer: $(5,-3)$

Question 8

(a) $y = x + 2$

The mirror line passes through the midpoint of $PP'$ : $\left(\frac{0 + 4}{2}, \frac{6 + 2}{2}\right) = (2, 4)$ .
The gradient of $PP'$ is $\frac{2 - 6}{4 - 0} = -1$ . The mirror line is perpendicular to $PP'$ , so its gradient is $1$ .
Through $(2, 4)$ with gradient 1:
$y - 4 = 1(x - 2) \;\Rightarrow\; y = x + 2$

P1 Finding the midpoint $(2, 4)$ .
P1 Using a gradient of 1, perpendicular to $PP'$ .
A1 $y = x + 2$ or any equivalent.

(b) Reflection in the line $y = -x$

Each vertex $(x, y)$ goes to $(-y, -x)$ , for example $(5, 3) \to (-3, -5)$ .
That is a reflection in the line $y = -x$ .

B1 Naming a reflection.
B1 The mirror line $y = -x$ .

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Reflections, rotations and translations

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