Pearson Edexcel International GCSE Mathematics A 4MA1: Vectors and transformation geometry – 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 cover topic 5, Vectors and transformation geometry, of the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), first assessed in June 2018 with papers in January and June. They test section 5.1 (Vectors) and section 5.2 (Transformation geometry). Questions marked (Higher tier only) use section 5.1, which is not on the Foundation papers. Questions 1 to 5 and 10 use section 5.2, which both tiers need. A calculator may be used on every 4MA1 paper, but show your working: coordinates written with no method can lose marks if one is wrong. Column vectors are written on one line as (x, y), top number first.

Related: study guide, revision notes, course hub, printable checklist.

Questions

1. Triangle A has vertices (1, 2), (3, 2) and (1, 5). Translate triangle A by the vector (−4, 3). Write down the coordinates of the vertices of the image. [2]

2. This question is about reflections.

(a) Triangle B has vertices (1, 3), (2, 5) and (4, 4). Reflect triangle B in the line y = x. Write down the coordinates of the image. [2] (b) A reflection maps the point (2, 5) onto the point (2, −1). Write down the equation of the mirror line. [1]

3. Triangle C has vertices (2, 1), (4, 1) and (4, 2). Rotate triangle C through −90° about the point (1, −1). Write down the coordinates of the vertices of the image. [3]

4. Triangle D has vertices (6, 5), (10, 5) and (6, 9).

(a) Enlarge triangle D by scale factor 1/2, centre (−2, 1). Write down the coordinates of the image. [3] (b) State one property of triangle D that does not change under this enlargement. [1]

5. Shape P has vertices (2, 1), (4, 1) and (2, 2). Shape Q has vertices (−3, 4), (−1, 4) and (−3, 5). Shape R has vertices (0, 1), (0, 3) and (−1, 1).

(a) Describe fully the single transformation that maps P onto Q. [2] (b) Describe fully the single transformation that maps P onto R. [3]

6. (Higher tier only) a = (−3, 4) and b = (5, −2).

(a) Find 3a + 2b as a column vector. [2] (b) Calculate |a − b|. [2]

7. (Higher tier only) Find the values of k and m such that k(2, −1) + m(1, 3) = (7, 7). [3]

8. (Higher tier only) ABCDEF is a regular hexagon. →AB = a and →BC = b.

(a) Explain why →DE = −a. [1] (b) You are given that →CD = b − a. Find →AD in terms of b. [2] (c) Find →AE in terms of a and b. [2]

9. (Higher tier only) OACB is a parallelogram. →OA = a and →OB = b. M is the midpoint of AC. P is the point on OM such that OP : PM = 2 : 1.

(a) Find →AB in terms of a and b. [1] (b) Find →OM in terms of a and b. [2] (c) Find →OP in terms of a and b. [1] (d) Prove that P lies on AB, and find the ratio AP : PB. [4]

10. Triangle T has vertices (−4, 2), (2, 2) and (−4, 6).

(a) Enlarge triangle T by scale factor 1/2, centre (0, 0). Label the image U and write down its vertices. [2] (b) Reflect triangle U in the line x = −1. Label the image V and write down its vertices. [2] (c) The longest side of T has length √52. Find the length of the longest side of V, giving a reason. [2]

11. (Higher tier only) The points A, B and C have coordinates A(−2, 1), B(4, −1) and C(6, 5).

(a) Write →AB and →BC as column vectors. [2] (b) Show that |→AB| = |→BC|. [2] (c) ABCD is a parallelogram. Find the coordinates of D. [2] (d) Show that a rotation through −90° about B maps A onto C. [3]

Answers

1. Add (−4, 3) to each vertex; at least two vertices correct [1]; all three correct: (−3, 5), (−1, 5), (−3, 8) [1] [2] Examiner insight: A translation by (−4, 3) applied as (+4, −3) or with the components swapped earns nothing; one slip in one vertex can still keep the first mark.

2. (a) Swap the coordinates; at least two vertices correct [1]; all three correct: (3, 1), (5, 2), (4, 4) [1] (b) The midpoint of (2, 5) and (2, −1) is (2, 2), and the segment is vertical, so the mirror line is horizontal: y = 2 [1] Examiner insight: The vertex (4, 4) lies on y = x, so it maps to itself; moving it is a sign the rule has been applied as a sign change rather than a swap. In (b), “x = 2” is a typical wrong line and scores zero.

3. Subtract the centre: (1, 2), (3, 2), (3, 3) [1]; turn −90° using (x, y) → (y, −x): (2, −1), (2, −3), (3, −3) [1]; add the centre back: (3, −2), (3, −4), (4, −4) [1] [3] Examiner insight: A rotation of +90° instead of −90°, or a correct turn about the origin instead of (1, −1), usually keeps at most one mark; check the image is the same size and shape as the object.

4. (a) Vectors from the centre: (8, 4), (12, 4), (8, 8) [1]; half of each: (4, 2), (6, 2), (4, 4) [1]; add the centre back: (2, 3), (4, 3), (2, 5) [1] (b) The angles (the size of each angle) stay the same [1] Examiner insight: Halving the coordinates themselves gives (3, 2.5), (5, 2.5), (3, 4.5), which is an enlargement about the origin and does not earn the final mark. In (b), “the shape” is too vague; name angles.

5. (a) Translation [1] by the vector (−5, 3) [1] (b) Rotation [1]; +90° (90° anticlockwise) [1]; centre (1, 0) [1] Check: (2, 1) − (1, 0) = (1, 1), turned +90° gives (−1, 1), and (1, 0) + (−1, 1) = (0, 1), the first vertex of R. Examiner insight: Each fact is a separate mark, so a description with no centre can score at most two out of three; giving two transformations (for example “reflection then translation”) scores zero.

6. (a) 3a = (−9, 12) and 2b = (10, −4) [1]; 3a + 2b = (1, 8) [1] (b) a − b = (−8, 6) [1]; |a − b| = √(64 + 36) = √100 = 10 [1] Examiner insight: Finding |a| − |b| (5 − √29) instead of |a − b| earns no marks; subtract the vectors first, then find the length.

7. Equate components: 2k + m = 7 and −k + 3m = 7 [1]; eliminate one unknown, for example m = 7 − 2k so −k + 21 − 6k = 7, giving 7k = 14 [1]; k = 2, m = 3 [1] [3] Examiner insight: Setting up both component equations earns the first mark even if the algebra later slips; check the answer by substituting back into both.

8. (a) In a regular hexagon DE is parallel to AB and the same length, but it points in the opposite direction, so →DE = −a [1] (b) →AD = →AB + →BC + →CD = a + b + (b − a) [1] = 2b [1] (c) →AE = →AD + →DE = 2b + (−a) [1] = 2b − a [1] Examiner insight: In (a), “same length” alone is not enough for the mark; the explanation must mention the opposite direction.

9. (a) →AB = −a + b = b − a [1] (b) →AC = →OB = b, so →AM = ½b [1]; →OM = →OA + →AM = a + ½b [1] (c) →OP = (2/3)→OM = (2/3)a + (1/3)b [1] (d) →AP = →OP − →OA = −(1/3)a + (1/3)b [1]; = (1/3)(b − a) [1]; so →AP = (1/3)→AB: AP is parallel to AB and A is a common point, so P lies on AB [1]; AP = (1/3)AB, so AP : PB = 1 : 2 [1] Examiner insight: The third mark in (d) needs the written conclusion naming both the multiple and the shared point A; correct algebra with no conclusion loses it.

10. (a) Halve each coordinate (centre at the origin); at least two correct [1]; all correct: U: (−2, 1), (1, 1), (−2, 3) [1] (b) Use (x, y) → (−2 − x, y); at least two correct [1]; all correct: V: (0, 1), (−3, 1), (0, 3) [1] (c) √13 (about 3.61) [1]; the enlargement with scale factor ½ halves every length, and a reflection keeps lengths the same [1] Examiner insight: Allow follow-through in (b) from a wrong U if the reflection is done correctly; in (c), the reason must mention both transformations.

11. (a) →AB = (4 − (−2), −1 − 1) = (6, −2) [1]; →BC = (2, 6) [1] (b) |→AB| = √(6² + (−2)²) = √40 [1]; |→BC| = √(2² + 6²) = √40, so the lengths are equal [1] (c) In parallelogram ABCD, →AD = →BC = (2, 6) [1]; D = (−2 + 2, 1 + 6) = (0, 7) [1] (d) →BA = (−6, 2) [1]; a −90° turn sends (x, y) to (y, −x), giving (2, 6) [1]; B + (2, 6) = (4 + 2, −1 + 6) = (6, 5), which is C [1] Examiner insight: Part (b) is a “show that”: both roots must be seen as √40 with a statement that they are equal, and decimals alone (6.32 and 6.32) with no exact working usually lose the final mark.

Where marks are usually lost

• Leaving out the centre of a rotation or enlargement, or giving it as a vector instead of a point.
• Giving an angle of rotation with no direction or sign.
• Describing a combination of transformations when one single transformation is asked for.
• Swapping x = a and y = b mirror lines.
• Enlarging from the origin when another centre is stated.
• Finding a translation vector as object − image instead of image − object.
• In vectors, writing →AB as a − b when →OA = a and →OB = b.
• In a proof, stopping at the algebra without stating “parallel” and the common point.
• Confusing |a − b| with |a| − |b|.
• Rounding a modulus early in a multi-step question.

Next steps

• Revision notes for this topic
• Study guide for this topic
• Edexcel IGCSE Mathematics hub
• Printable 4MA1 checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Pearson Edexcel International GCSE in Mathematics (Specification A) (4MA1), Specification Issue 2, November 2017, Pearson Education Limited. Topic 5, Vectors and transformation geometry: 5.1 Vectors (Higher tier only) and 5.2 Transformation geometry.