Transformations & Vectors: IGCSE Mathematics questions and answers
10 free IGCSE Mathematics Transformations & Vectors questions written in the style of Cambridge Mathematics papers, each with its answer and mark scheme. 7 of them are A* challenge questions, built to catch the mistakes strong students still make.
This topic covers Transformations, Vectors in two dimensions, Magnitude of a vector. Work through the questions below, then generate a full Transformations & Vectors paper with a fresh set every time.
In triangle OAB, M is the midpoint of OA and N is the midpoint of OB. Prove that MN is parallel to AB and half its length.
[4]
(b)
PQRS is any quadrilateral. Prove that the midpoints of its four sides are the vertices of a parallelogram.
[5]
(c)
Prove that this parallelogram is a rhombus if and only if the diagonals PR and QS of the original quadrilateral are equal in length.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
MN = ½b − ½a = ½(b − a) = ½AB
OA = a, OB = b, AB = b − a B1 MN = ½b − ½a M1 = ½AB A1 so parallel and half the length A1
4
(b)
each pair of opposite sides of the midpoint quadrilateral equals ½PR or ½QS
position vectors p, q, r, s and midpoints ½(p + q) etc. M1 side joining midpoints of PQ and QR = ½(r − p) A1 side joining midpoints of RS and SP = ½(p − r) A1 opposite sides equal and parallel (same for the other pair) M1 so a parallelogram A1
5
(c)
adjacent sides are ½PR and ½QS
adjacent sides of the parallelogram are ½(r − p) and ½(s − q) M1 rhombus ⇔ these lengths are equal ⇔ PR = QS A1 both directions argued A1
3
Question 2Structured · 11 marksA* challenge
Two ships, A and B, sail at constant velocities. Their position vectors relative to a harbour O, t hours after noon, are OA = 29 + t4−1 and OB = −63 + t61, where distances are in kilometres and 10 and 01 represent 1 km east and 1 km north.
(a)
Find the least distance between the ships and the time at which this occurs.
[5]
(b)
The ships can communicate by radio when they are within 5 km of each other. Find the length of time for which they can communicate. Give your answer in hours and minutes, correct to the nearest minute.
[4]
(c)
Find the bearing of ship B from ship A at the moment when they are closest together.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
√2 km (1.41 km) at 15 30
AB = −8 + 2t−6 + 2tM1 AB2 = (2t − 8)2 + (2t − 6)2 = 8t2 − 56t + 100 A1 completing the square or differentiating: t = 3.5 M1 least AB2 = 2, distance √2 = 1.41 km A1 15 30 (3.30 pm) A1
5
(b)
3 hours 23 minutes
8t2 − 56t + 100 ≤ 25 M1 8t2 − 56t + 75 = 0, formula M1 t = 1.804 and 5.196 (or difference √462 = 3.391 h) A1 3 h 23 min A1
4
(c)
315°
AB = −11 at t = 3.5 M1 315° A1
2
Question 3Structured · 12 marksA* challenge
E1 is the enlargement with centre (0, 0) and scale factor 2. E2 is the enlargement with centre (6, 0) and scale factor −12. Triangle T has vertices (1, 1), (3, 1) and (1, 4).
(a)
T is transformed by E1 and then the image is transformed by E2. Find the coordinates of the vertices of the final image, and describe fully the single transformation that maps T onto this final image.
[5]
(b)
E3 is an enlargement with scale factor k and centre (c, 0). E1 followed by E3 is the translation by the vector 60. Find the values of k and c.
[3]
(c)
The enlargement with centre (0, 0) and scale factor 3, followed by the enlargement with centre (a, b) and scale factor 13, is the translation by the vector 4−2. Find the values of a and b.
k = 12 (sizes must be unchanged) B1 (x, y) → (c + ½(2x − c), y) = (x + c2, y) M1 c = 12 A1
3
(c)
a = 6, b = −3
image of (x, y): (a + 13(3x − a), b + 13(3y − b)) M1 = (x + 2a3, y + 2b3) A1 2a3 = 4 and 2b3 = −2 M1 a = 6, b = −3 A1
4
Question 4Structured · 8 marksA* challenge
OAB is a triangle with OA = a and OB = b. P is the point on AB such that AP : PB = 2 : 1. Q is the point on OB such that OQ = 23b. The lines OP and AQ intersect at X.
(a)
Find OP in terms of a and b.
[2]
(b)
By writing OX = λOP and AX = μAQ, find the values of λ and μ.
[5]
(c)
Find the ratio OX : XP in its simplest form.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
13a + 23b
a + 23(b − a) M1 13a + 23b A1
2
(b)
λ = 34, μ = 34
OX = a + μ(−a + 23b) M1 equate coefficients of a and of b M1 two correct equations A1 λ = 34A1 μ = 34A1
5
(c)
3 : 1
3 : 1 B1
1
Question 5Structured · 4 marks
a = −44 b = −4−1
(a)
Find 4a − b.
[2]
(b)
Find |a + b|.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
−1217
One component correct B1 −1217B1
2
(b)
8.54
√((−8)2 + 32) M1 8.54 oe A1
2
Question 6Structured · 5 marks
The diagram shows shape A and shape B on a grid.
Fig. 6.1
(a)
Describe fully the single transformation that maps shape A onto shape B.
[3]
(b)
On the grid, reflect shape A in the line y = 3. Label the image C.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
Rotation, 180°, centre (−2, −1)
Rotation B1 180° B1 centre (−2, −1) B1
3
(b)
Image C drawn with vertices (−4, 6), (−2, 6), (−2, 5), (−4, 4)
Correct image B2 B1 for reflection in another horizontal line or correct reflection of the line y = 3 with one vertex wrong
2
Question 7Structured · 3 marks
p = −3−2
(a)
Find |p|.
[2]
(b)
Find 3p.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
3.61
√((−3)2 + (−2)2) M1 3.61 A1
2
(b)
−9−6
−9−6B1
1
Question 8Structured · 7 marksA* challenge
OAB is a triangle with OA = a and OB = b. P is the point on AB such that AP : PB = 1 : 3. M is the midpoint of OA. The point R is such that OR = 3OP.
(a)
Find, in terms of a and b, in their simplest form, the vector OP.
[2]
(b)
Find MR in terms of a and b.
[2]
(c)
The line BR meets the line OA, extended, at the point X, where OX = ta. Find the value of t.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
34a + 14b
OP = a + 14(b − a) M1 34a + 14b A1
2
(b)
74a + 34b
MR = MO + OR M1 74a + 34b A1
2
(c)
t = 9
BR = −b + OR, and OX = OB + μBR M1 coefficient of b equal to zero: μ = 4 M1 t = 9 A1
3
Question 9Structured · 9 marksA* challenge
O is the origin. The position vector of A is a and the position vector of B is b. P is the point on AB such that AP : PB = 2 : 1. M is the midpoint of OA. R is the point with position vector 2b.
Find, in terms of a and/or b, in its simplest form, the vector
(a)(i)
from A to B,
[1]
(ii)
from O to P,
[2]
(iii)
from M to P.
[2]
(b)
Show that M, P and R lie on a straight line, and find the ratio MP : PR.
[4]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)(i)
b − a
b − a B1
1
(ii)
13a + 23b
a + 23(b − a) M1 13a + 23b A1
2
(iii)
−16a + 23b
(vector OP) − ½a M1 −16a + 23b A1
2
(b)
vector MR = −½a + 2b = 3 × (vector MP); parallel with common point M, so collinear. MP : PR = 1 : 2
vector MR = −½a + 2b B1 vector MR = 3 × vector MP M1 parallel and share point M, so collinear A1 MP : PR = 1 : 2 B1
4
Question 10Structured · 6 marksA* challenge
OABC is a parallelogram. O is the origin, OA = a and OC = c. M is the midpoint of AB.
(a)
Find AB in terms of c.
[1]
(b)
Find OM in terms of a and c.
[2]
(c)
N is the point on OB such that ON : NB = 2 : 1. Find MN in terms of a and c.
What is in IGCSE Mathematics Transformations & Vectors?
In the 0580 syllabus, Transformations & Vectors covers Transformations, Vectors in two dimensions, Magnitude of a vector.
Where can I get Transformations & Vectors questions with mark schemes?
This page has 10 Transformations & Vectors questions with answers. For more, the free generator on mentorriyaz.in builds a new Transformations & Vectors paper with a full mark scheme each time you click.
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in