OCR GCSE Mathematics: Isometric transformations and vectors (J560) – Practice Questions

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 follow OCR GCSE (9-1) Mathematics, specification Version 2.1 (August 2026), sections 9.01a to 9.01d and 9.03a and 9.03b. Questions 1 to 7 are for both tiers. Questions 8 to 12 are marked (Higher tier only). In this set a column vector is written (top over bottom).

Revise first with the study guide or the revision notes. Use the checklist and the course hub.

Questions

1. Triangle PQR has vertices P(1, −2), Q(4, −2) and R(1, −4). Write the coordinates of its image after reflection in the line x = −1. [3]

2. Triangle T has vertices (3, 1), (5, 1) and (5, 4). Its image has vertices (1, 3), (1, 5) and (4, 5).

(a) Give the equation of the mirror line. [2] (b) The point (−2, 6) is reflected in the line y = −x. Find its image. [1]

3. A triangle has vertices (3, 1), (6, 1) and (6, 3). Rotate it through 90° anticlockwise about (2, 0). Write the coordinates of the image. [3]

4. Shape F has vertices (1, 1), (4, 1) and (4, 2). Its image has vertices (2, −4), (2, −7) and (3, −7). Describe fully the single transformation. [4]

5. The vertex (2, 5) of a tile maps to (−1, 3) under a translation.

(a) Write the translation as a column vector. [1] (b) Write the translation that maps the image back onto the tile. [1] (c) Find the image of (4, −1) under the original translation. [1]

6. a = (3 over −2) and b = (−4 over 1).

(a) Work out 2a + 3b. [2] (b) Work out a − 2b. [2] (c) The vector (9 over y) is a multiple of a. Find y. [1]

7. P(−2, 1), Q(3, 4) and R(7, 2) are three vertices of parallelogram PQRS.

(a) Write PQ and QR as column vectors. [2] (b) Find the coordinates of S. [2]

8. (Higher tier only) Shape D is reflected in the horizontal line y = −2. Its image is then reflected in the horizontal line y = 5.

(a) Describe the single transformation that maps D onto the final image. [3] (b) Explain why no point is invariant. [1]

9. (Higher tier only) Shape S is reflected in the x-axis and then in the line y = x.

(a) Describe the single transformation that has the same effect. [3] (b) The reflections are done in the opposite order. Describe the single transformation. [2]

10. (Higher tier only) A shape is rotated through 90° anticlockwise about the origin, then translated by (2 over 4).

(a) Find the final image of the point (3, 1). [2] (b) Show that the point (−1, 3) is invariant under the combined transformation. [2] (c) Describe the single transformation. [2]

11. (Higher tier only) In quadrilateral ABCD, AB = 2p + q, BC = p − 3q and CD = −6p − 3q. Prove that ABCD is a trapezium. [5]

12. (Higher tier only) ORS is a triangle with OR = r and OS = s. The point T is on RS with RT : TS = 2 : 3. The point U is such that OU = 3r + 2s.

(a) Show that OT = (1/5)(3r + 2s). [3] (b) Prove that O, T and U are collinear. [2] (c) Find the ratio OT : TU. [1]

Answers

1. Use x → 2(−1) − x = −2 − x. P(1, −2) goes to (−3, −2) [1]. Q(4, −2) goes to (−6, −2) [1]. R(1, −4) goes to (−3, −4) [1]. Image: (−3, −2), (−6, −2), (−3, −4) Examiner insight: Give all three image vertices as coordinates, and count from the mirror line, not from the y-axis. Check each distance to the line is equal on both sides.

2. (a) (3, 1) goes to (1, 3): the coordinates swap [1]. The mirror line is y = x [1]. (b) (−y, −x) gives (−6, 2) [1]. Examiner insight: Test the line on one more vertex. Writing “reflection” without the equation does not answer a question that asks for the equation of the mirror line.

3. Relative to (2, 0): (1, 1), (4, 1), (4, 3). Turn anticlockwise (−dy, dx): (−1, 1), (−1, 4), (−3, 4) [1]. Add the centre back: (1, 1), (1, 4), (−1, 4) [1]. Image: (1, 1), (1, 4), (−1, 4) [1] Examiner insight: Show the relative positions so that each step of the turn can be followed and checked.

4. Try the centre (−1, −2). Its squared distance from both (1, 1) and (2, −4) is 13 [1]. From the centre, (1, 1) is at (2, 3) and its image is at (3, −2) [1]. Turning (2, 3) a quarter turn clockwise gives (3, −2) [1]. A rotation of 90° clockwise about (−1, −2) [1] Examiner insight: A rotation answer needs three items: centre, angle and sense. Naming only “rotation 90°” leaves the answer incomplete.

5. (a) (−1 − 2 over 3 − 5) = (−3 over −2) [1]. (b) (3 over 2) [1]. (c) (4 − 3, −1 − 2) = (1, −3) [1]. Examiner insight: Subtract object from image, in that order. Give a column vector rather than “3 left and 2 down” when the question says to write one.

6. (a) 2a = (6 over −4) and 3b = (−12 over 3) [1]. Sum: (−6 over −1) [1]. (b) 2b = (−8 over 2) [1]. a − 2b = (3 + 8 over −2 − 2) = (11 over −4) [1]. (c) 9 = 3 × 3, so y = 3 × (−2) = −6 [1]. Examiner insight: Show each scalar multiple before combining. Subtracting a negative number is a common slip, so write the sign change explicitly.

7. (a) PQ = (3 + 2 over 4 − 1) = (5 over 3) [1]. QR = (7 − 3 over 2 − 4) = (4 over −2) [1]. (b) In a parallelogram PS = QR [1]. S = (−2 + 4, 1 − 2) = (2, −1) [1]. Examiner insight: Say which equal vectors you use. Check by confirming SR equals PQ: both are (5 over 3).

8. (a) (3, 4) goes to (3, −8), then to (3, 18) [1]. The shift is 14 upwards, and 14 = 2 × (5 − (−2)) [1]. A translation by (0 over 14) [1]. (b) A translation moves every point, so none stays fixed [1]. Examiner insight: Name the transformation type first, then give its full details. Tracking one point lets you check the shortcut for parallel mirror lines.

9. (a) Take (2, 1). The x-axis sends it to (2, −1) [1], then y = x sends it to (−1, 2) [1]. This is a quarter turn anticlockwise about the origin where the mirror lines meet: rotation 90° anticlockwise about (0, 0) [1]. (b) (2, 1) goes to (1, 2), then to (1, −2) [1]. Rotation 90° clockwise about (0, 0) [1]. Examiner insight: Use the same test point for both orders. Showing that the results differ is the clearest way to show that order matters.

10. (a) Rotation sends (x, y) to (−y, x), so (3, 1) goes to (−1, 3) [1]. Adding (2 over 4) gives (1, 7) [1]. (b) (−1, 3) goes to (−3, −1) under the rotation [1]. Adding (2 over 4) gives (−1, 3), the starting point, so it is invariant [1]. (c) The invariant point is the centre. Check: relative to (−1, 3), the point (3, 1) is at (4, −2) and its image (1, 7) is at (2, 4), which is a 90° anticlockwise turn [1]. Rotation of 90° anticlockwise about (−1, 3) [1] Examiner insight: In a “show that” part, write the start point, the point after each step and the final point. An unsupported statement that the point is invariant is hard to credit.

11. DC = −CD = 6p + 3q [1]. 6p + 3q = 3(2p + q) = 3AB [1]. So DC is parallel to AB [1]. AD = AB + BC + CD = −3p − 5q [1]. BC = p − 3q is not a multiple of AD, so BC and AD are not parallel and ABCD has exactly one pair of parallel sides: a trapezium [1] Examiner insight: Finish with a sentence that links the algebra to the geometry. Showing the second pair is not parallel supports “trapezium” rather than “parallelogram”.

12. (a) RS = s − r [1]. RT = (2/5)RS, since RT : TS = 2 : 3 [1]. OT = r + (2/5)(s − r) = (3/5)r + (2/5)s = (1/5)(3r + 2s) [1]. (b) OU = 3r + 2s = 5OT [1]. The vectors share the point O and are multiples, so O, T and U are on one line [1]. (c) TU = OU − OT = 4OT, so OT : TU = 1 : 4 [1]. Examiner insight: A “show that” question needs each line of the route written down, ending on the stated result. Give the reason (shared point plus multiple), not just the multiple.

Where marks are usually lost

• Giving a rotation or reflection without every detail asked for.
• Mixing up clockwise and anticlockwise, and not checking with a point.
• Measuring a reflection from an axis when the mirror line is x = a or y = b.
• Writing a translation as coordinates or words, not a column vector.
• Subtracting in the wrong order when finding a vector from two points.
• Slipping on signs when subtracting a vector with negative entries.
• Applying two transformations in the wrong order, or applying the second to the original shape.
• Stating that vectors are parallel without showing the multiple.
• Concluding a proof without a sentence naming the geometric result.

Next steps

• Re-read the revision notes and the study guide.
• Go to the OCR GCSE Mathematics hub and the checklist.
• Take the free 10-minute diagnostics to find your next topic.
• Book a free trial class: /trial/.

Official syllabus

OCR GCSE (9-1) Mathematics (J560), Version 2.1 (August 2026), for first assessment in 2017, published by OCR. This set covers sections 9.01a to 9.01d and 9.03a and 9.03b.