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Cambridge O Level Mathematics: Transformations: Reflection, Rotation, Enlargement and Translation (4024) – Practice Questions

Original practice questions with worked answers on reflection, rotation, enlargement and translation for Cambridge O Level Mathematics 4024.

Subject
Mathematics
Level
O LEVELS
Topic
Transformations: Reflection, Rotation, Enlargement and Translation
Updated

Aligned to Cambridge O Level Mathematics (4024), 2025-2027. Official specification .

Syllabus page (what it covers and how it is assessed): Cambridge O Level Mathematics.

Syllabus points this page covers

4024

  • 7.1 Transformations

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These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 7.1 (Transformations) of the Cambridge O Level Mathematics (Syllabus D) 4024 syllabus for 2025, 2026 and 2027, which is not tiered. Learn the methods first in the study guide or the revision notes. Sketch every question on squared paper and use a ruler for straight edges. The course hub, the printable checklist and the free diagnostics show what to do next.

Questions

1. Point P is (6, −4). Write the coordinates of the image of P after (a) reflection in the line x = −1 [1] (b) reflection in the line y = −x [1] (c) rotation through 90° anticlockwise about the origin. [1]

2. Triangle L has vertices (−5, −1), (−2, −1) and (−2, −3). Find the vertices of the image of L after rotation through 180° about the point (−1, 1). [3]

3. Triangle S has vertices (2, 3), (6, 3) and (6, 5). Its image T has vertices (2, −1), (6, −1) and (6, −3). Describe fully the single transformation that maps S onto T. [2]

4. Triangle R has vertices (4, 2), (8, 2) and (8, 6) and area 8 square units. (a) Find the vertices of the image of R after an enlargement with scale factor −1/2 and centre (2, 0). [3] (b) Write down the area of the image. [1]

5. Triangle F has vertices (7, −2), (11, −2) and (11, 2). Its image G has vertices (5, −1), (7, −1) and (7, 1). Describe fully the single transformation that maps F onto G. [4]

6. A translation maps A(−2, 5) onto A′(4, 1). (a) Write down the column vector of the translation. [1] (b) Find the image of B(0, −3). [1] (c) Find the point whose image is (1, 1). [1]

7. Shape M is reflected in the line x = 2. The image is then reflected in the line x = 7. (a) Describe fully the single transformation that maps M onto its final image. [3] (b) A vertex of M is at (3, 1). Find its final position. [1]

8. Triangle K has vertices (1, 2), (1, 5) and (3, 2). K is rotated through 90° clockwise about the origin to give K₁. K₁ is then reflected in the y-axis to give K₂. (a) Find the vertices of K₁. [2] (b) Find the vertices of K₂. [1] (c) Describe fully the single transformation that maps K onto K₂. [2]

9. (a) A photograph has area 12 cm². It is enlarged with scale factor 2.5. Find the area of the enlargement. [2] (b) An enlargement maps a triangle of area 45 cm² onto a triangle of area 5 cm². Find the possible scale factors. [2] (c) An enlargement has scale factor −3 and centre C. Point P is 4 cm from C. Describe where the image of P lies. [2]

10. Triangle A has vertices (−6, 2), (−3, 2) and (−3, 4). (a) A is rotated through 90° clockwise about (−3, 2) to give B. Find the vertices of B. [3] (b) B is enlarged with scale factor −2 and centre (−3, 2) to give C. Find the vertices of C. [3] (c) Find the area of C and show it is 4 times the area of A. [2]

11. Triangle X has vertices (1, 1), (4, 1) and (4, 3). Triangle Y has vertices (0, 4), (0, 7) and (−2, 7). Triangle Z has vertices (4, 0), (4, 3) and (2, 3). (a) Describe fully the single transformation that maps X onto Y. [4] (b) Describe fully the single transformation that maps Y onto Z. [1] (c) Describe fully the single transformation that maps X onto Z. [3]

Answers

1. (a) Image x = 2(−1) − 6 = −8, so (−8, −4) [1]. (b) (x, y) → (−y, −x) gives (4, −6) [1]. (c) (x, y) → (−y, x) gives (4, 6) [1]. [3] Examiner insight: Write the rule you are using before you substitute. A slip in one part is then easy to see and does not spread.

2. For a half turn about (−1, 1), image = (−2 − x, 2 − y) [1]. (−5, −1) → (3, 3); (−2, −1) → (0, 3); (−2, −3) → (0, 5) [2]. [3] Examiner insight: A method shown for one vertex lets an examiner see what you intended if a later vertex is wrong. Sketch the triangle to check that the image is the right way round.

3. The x values are unchanged and the y values change, so it is a reflection [1] in the line y = 1 [1], halfway between 3 and −1. [2] Examiner insight: Give the mirror line as an equation. “Horizontal line through the middle” is not a full description.

4. (a) Positions relative to (2, 0): (2, 2), (6, 2), (6, 6) [1]. Multiply by −1/2: (−1, −1), (−3, −1), (−3, −3) [1]. Add (2, 0): (1, −1), (−1, −1), (−1, −3) [1]. (b) Area × (−1/2)² = 8 × 1/4 = 2 square units [1]. [4] Examiner insight: Show the relative positions. A negative fraction can easily send an image to the wrong side of the centre, and working lets the method be seen.

5. The image is smaller, so it is an enlargement [1]. Side FG is 4 and its image is 2, with the shape the same way up, so scale factor 1/2 [1]. Draw lines through F and its image, and through G and its image [1]. They cross at centre (3, 0) [1]. [4] Examiner insight: A description of an enlargement needs both scale factor and centre. Lines drawn with a ruler through matching points make the centre easy to read.

6. (a) (4 − (−2), 1 − 5) = (6 over −4) [1]. (b) (0 + 6, −3 − 4) = (6, −7) [1]. (c) Subtract the vector: (1 − 6, 1 + 4) = (−5, 5) [1]. [3] Examiner insight: Part (c) works backwards. Check by adding the vector to your answer to see that it returns (1, 1).

7. (a) Two reflections in parallel lines give a translation [1]. The distance moved is 2 × (7 − 2) = 10 [1], in the direction perpendicular to the lines: (10 over 0) [1]. (b) (3, 1) → (1, 1) → (13, 1) [1]. [4] Examiner insight: Doing part (b) point by point is a useful check on your answer to part (a).

8. (a) (x, y) → (y, −x): (2, −1), (5, −1), (2, −3) [2]. (b) Change the sign of each x value: (−2, −1), (−5, −1), (−2, −3) [1]. (c) K(1, 2) maps to (−2, −1), which fits (x, y) → (−y, −x) [1]. This is a reflection in the line y = −x [1]. [5] Examiner insight: To name a combination, follow one or two points through, then match the result to a rule or a picture.

9. (a) Area scale factor 2.5² = 6.25 [1]; 12 × 6.25 = 75 cm² [1]. (b) Area ratio 5 ÷ 45 = 1/9 [1], so the scale factor is 1/3 or −1/3 [1]. (c) CP′ = 3 × 4 = 12 cm [1], on the opposite side of C from P, on the straight line through P and C [1]. [6] Examiner insight: Area uses k², so a root gives two answers. Say that the negative value is allowed because the question does not rule it out.

10. (a) Positions relative to (−3, 2): (−3, 0), (0, 0), (0, 2) [1]. Clockwise (u, v) → (v, −u): (0, 3), (0, 0), (2, 0) [1]. Add (−3, 2): (−3, 5), (−3, 2), (−1, 2) [1]. (b) Multiply B’s relative positions (0, 3), (0, 0), (2, 0) by −2: (0, −6), (0, 0), (−4, 0) [1]. Add (−3, 2): (−3, −4), (−3, 2), (−7, 2) [2]. (c) Area of A = 1/2 × 3 × 2 = 3, area of C = 1/2 × 6 × 4 = 12 [1]. 12 = 4 × 3, which matches (−2)² [1]. [8] Examiner insight: Keep the centre fixed in your head. It is invariant under both transformations here, which is a quick check on your vertices.

11. (a) It is a rotation [1]. Side (1, 1) to (4, 1) points right and its image (0, 4) to (0, 7) points up, so 90° anticlockwise [1]. The perpendicular bisector of (1, 1) to (0, 4) is x = 3y − 7, and that of (4, 1) to (0, 7) is 3y = 2x + 8 [1]. They meet at centre (−1, 2) [1]. (b) (0, 4) → (4, 0), so it is a translation by (4 over −4) [1]. (c) The sides turn through a quarter turn anticlockwise, so it is a rotation [1] of 90° anticlockwise [1] about the point where the perpendicular bisectors of (4, 1) to (4, 3) and of (4, 3) to (2, 3) meet, y = 2 and x = 3: centre (3, 2) [1]. [8] Examiner insight: A rotation followed by a translation is again a rotation of the same angle, with a new centre. Always test the centre you find on a second vertex.

Where marks are usually lost

  • Omitting the point a rotation or enlargement is about, or the direction of a turn.
  • Describing a reflection as “a flip” or giving the mirror line in words only.
  • Placing an enlargement from the shape’s middle instead of the stated centre.
  • Using a negative scale factor and keeping the image on the same side of the centre.
  • Using the scale factor for areas without squaring it.
  • Combining two transformations in the wrong order.
  • Drawing sloping edges freehand, when a ruler must be used.
  • Writing a translation as a coordinate pair when a column vector is asked for.

Next steps

Official syllabus

Cambridge O Level Mathematics (Syllabus D) 4024 syllabus for 2025, 2026 and 2027, published by Cambridge University Press & Assessment (Cambridge International), section 7.1 Transformations.

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