Worksheets · Higher

Combined transformations and invariance

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

Question 1Non-calculator · 3 marks
Shape $S$ is reflected in the line $y = x$ to give shape $S_1$ . $S_1$ is then reflected in the $x$ -axis to give shape $S_2$ .
(a) Describe fully the single transformation that maps $S$ onto $S_2$ . (3)
Question 2Non-calculator · 3 marks
(a) Point P( $-2$ , 4) is reflected in the line x = 1. Its image is then reflected in the line x = 5. Work out the coordinates of the final image. (3)
Question 3Non-calculator · 3 marks
Triangle $P$ has vertices $(1, 1)$ , $(4, 1)$ and $(4, 3)$ .
(a) $P$ is reflected in the line $y = x$ . Write down the coordinates of the vertex that is invariant. (1)
(b) $P$ is rotated $180^\circ$ about the point $(4, 2)$ . Explain why no vertex of $P$ is invariant, but one point of the triangle is. (2)
Question 4Non-calculator · 4 marks
A point P $(2,-3)$ is reflected in the line x = 3, then in the line x = 6.
(a) Find the final image coordinates. (3)
(b) Describe the single transformation equivalent to the two reflections. (1)
Question 5Non-calculator · 5 marks
A point P $(3,1)$ is rotated 90 degrees clockwise about the origin and then translated by $(3,-2)$ .
(a) Find the final image of P. (3)
(b) Find the final image if the order is reversed. (2)
Question 6Non-calculator · 4 marks
A shape is reflected in the x-axis and then rotated 180 degrees about the origin.
(a) Find the final image of $(4,-4)$ . (3)
(b) Describe the equivalent single transformation. (1)
Question 7Non-calculator · 3 marks
Transformation A is a rotation of 90° clockwise about (0, 0). Transformation B is a reflection in the line y = -x.
(a) Find the image of (2, 5) under A followed by B. (1)
(b) Find the image of (2, 5) under B followed by A. (1)
(c) Describe fully the single transformation equivalent to A followed by B. (1)
Question 8Non-calculator · 3 marks
A shape is reflected in the line $x = a$ and then its image is reflected in the line $x = b$ .
(a) Prove that the combined transformation is a translation by the vector with components $2(b - a)$ and 0. (3)

Worked solutions and marks

Question 1

(a) Rotation $90^\circ$ clockwise about $(0, 0)$

Follow a general point: $(x, y) \to (y, x)$ after the first reflection.
Then $(y, x) \to (y, -x)$ after reflecting in the $x$ -axis.
$(x, y) \to (y, -x)$ is a rotation of $90^\circ$ clockwise about the origin. Check: $(1, 0) \to (0, -1)$ .

M1 Tracking a point through both reflections, reaching $(y, -x)$ or a correct numerical example such as $(1, 2) \to (2, -1)$ .
A1 Rotation, $90^\circ$ clockwise.
B1 Centre $(0, 0)$ .

Question 2

(a) (6, 4)

$2\times 1-(-2)$
$2\times 5-4$
Reflection in x = a sends x to 2a $-$ x and leaves y unchanged.
The first image is (2 $\times$ 1 $-$ ( $-2$ ), 4) = (4, 4).
The final image is (2 $\times$ 5 $-$ 4, 4) = (6, 4).

P1 Establishing $2\times 1-(-2)$ or an equivalent valid method.
P1 Establishing $2\times 5-4$ or an equivalent valid method.
C1 Correct conclusion with the complete supporting argument: (6, 4)

Question 3

(a) $(1, 1)$

A point is invariant under a reflection when it lies on the mirror line. Only $(1, 1)$ is on $y = x$ .

B1 Correct answer: $(1, 1)$ .

(b) Only the centre $(4, 2)$ is invariant under a rotation; it lies on the side from $(4, 1)$ to $(4, 3)$ but is not a vertex.

A rotation fixes only its centre.
$(4, 2)$ is the midpoint of the side joining $(4, 1)$ and $(4, 3)$ , so that point of the triangle stays put while every vertex moves.

C1 Only the centre of rotation is invariant.
C1 The centre $(4, 2)$ lies on a side of $P$ but is not a vertex.

Question 4

(a) $(8,-3)$

The first reflection replaces x by twice the mirror coordinate minus x.
$2\times 3-2$
Apply the same rule at the second mirror.
$2\times 6-(2\times 3-2)$
Therefore $(8,-3)$ .

M1 The first reflection replaces x by twice the mirror coordinate minus x.
M1 Apply the same rule at the second mirror.
A1 Correct answer: $(8,-3)$

(b) A translation 6 units to the right. Parallel mirrors three units apart give twice that separation in the direction from the first mirror to the second.

A translation 6 units to the right. Parallel mirrors three units apart give twice that separation in the direction from the first mirror to the second.

C1 Correct conclusion with supporting reasoning: A translation 6 units to the right. Parallel mirrors three units apart give twice that separation in the direction from the first mirror to the second.

Question 5

(a) $(4,-5)$

A clockwise quarter turn maps (x,y) to (y,-x).
$y=-3$
Add the translation after the rotation.
$1+3$
Therefore $(4,-5)$ .

M1 A clockwise quarter turn maps (x,y) to (y,-x).
M1 Add the translation after the rotation.
A1 Correct answer: $(4,-5)$

(b) $(-1,-6)$

Translate first to (6, −1), then apply the rotation rule.
$y=-(3+3)$
Therefore $(-1,-6)$ .

M1 Translate first to (6, −1), then apply the rotation rule.
A1 Correct answer: $(-1,-6)$

Question 6

(a) $(-4,-4)$

First negate only the y-coordinate.
$y=4$
Then negate both coordinates.
$x=-4$
Therefore $(-4,-4)$ .

M1 First negate only the y-coordinate.
M1 Then negate both coordinates.
A1 Correct answer: $(-4,-4)$

(b) Reflection in the y-axis: the combined rule is (x,y) to (-x,y).

Reflection in the y-axis: the combined rule is (x,y) to (-x,y).

C1 Correct conclusion with supporting reasoning: Reflection in the y-axis: the combined rule is (x,y) to (-x,y).

Question 7

(a) $(2,-5)$

Therefore $(2,-5)$ .

B1 Correct answer: $(2,-5)$

(b) $(-2,5)$

Therefore $(-2,5)$ .

B1 Correct answer: $(-2,5)$

(c) A reflection in the x-axis.

A reflection in the x-axis.

C1 Correct conclusion with supporting reasoning: A reflection in the x-axis.

Question 8

(a) $(x, y) \to (2a - x, y) \to (2b - 2a + x, y)$

Reflecting in $x = a$ keeps $y$ and sends $x$ to the same distance on the other side: $x \to 2a - x$ .
Reflecting that in $x = b$ : $2a - x \to 2b - (2a - x) = x + 2(b - a)$ .
So $(x, y) \to (x + 2(b - a),\; y)$ for every point: a translation by $2(b - a)$ across and 0 up.

M1 The first reflection written generally: $x \to 2a - x$ .
M1 The second reflection applied to $2a - x$ .
A1 Reaching $x + 2(b - a)$ with $y$ unchanged and concluding it is a translation.

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Combined transformations and invariance

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