ExaminerPrepexaminerprep.com · Vectors and transformation geometry Questions
Free consultation
Free access until 30 November 2026 · Physics and Maths questions and notes, updated continuously · Build a test

ExaminerPrep › Question bank › IGCSE Maths (Edexcel) › Unit 5 Vectors and transformation geometry

International GCSE Maths · Pearson Edexcel 4MA1 · Unit 5

Vectors and transformation geometry

Original exam-style questions with full mark schemes. Use the filters to choose level and difficulty, and tick questions to build a printable test.

26questions
123marks
Fullmark schemes
Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 15 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →

Make your own test. Tap + Add to test on any question, from any topic, then “Build & print test”. Or let the test builder choose questions for you in seconds.
1IGME-062-01-02Transformations described in words Paper FEasy4 marks

The point P has coordinates (5, 3).

(a)

P is reflected in the y-axis. Write down the coordinates of its image.

(1)
(b)

P is reflected in the line y = x. Write down the coordinates of its image.

(1)
(c)

P is reflected in the line x = 2. Find the coordinates of its image.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 (−5, 3)✓ 1
Part (b)
B1 (3, 5)✓ 1
Part (c)
M1 for using the fact that P is 3 units to the right of the line x = 2, so its image is 3 units to the left✓ 1
A1 (−1, 3)✓ 1

Common error: The line x = 2 is a vertical line, not a horizontal one. Reflecting in it changes only the x-coordinate, and the y-coordinate stays the same.

2IGME-062-01-03Transformations described in words Paper FEasy3 marks

The point Q has coordinates (4, 1).

(a)

Q is rotated 90° clockwise about the origin. Write down the coordinates of its image.

(1)
(b)

Q is rotated 180° about the origin. Write down the coordinates of its image.

(1)
(c)

Q is rotated 270° clockwise about the origin. Write down the coordinates of its image.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 (1, −4)✓ 1
Part (b)
B1 (−4, −1)✓ 1
Part (c)
B1 (−1, 4)✓ 1

Exam tip: A rotation of 270° clockwise is the same as a rotation of 90° anticlockwise. Use whichever turn is easier to picture.

3IGME-062-01-04Transformations described in words Paper FEasy4 marks

Triangle A has vertices at (1, 1), (2, 1) and (1, 3).
Triangle A is enlarged with scale factor 3 and centre (0, 0) to give triangle B.

(a)

Write down the coordinates of the vertices of triangle B.

(2)
(b)

Find the area of triangle B.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B2 for (3, 3), (6, 3) and (3, 9)✓ 2
(B1 for two correct vertices)—
Part (b)
M1 for area of A = ½ × 1 × 2 = 1, then × 32; or ½ × 3 × 6✓ 1
A1 9 square units✓ 1

Common error: An enlargement with scale factor 3 multiplies every length by 3. It multiplies the area by 3² = 9, not by 3.

4IGME-062-01-05Transformations described in words Paper FEasy5 marks

Shape A has vertices at (2, 1), (5, 1) and (5, 3).
Shape B has vertices at (2, −1), (5, −1) and (5, −3).
Shape C has vertices at (−2, 1), (−5, 1) and (−5, 3).

(a)

Describe fully the single transformation that maps shape A onto shape B.

(2)
(b)

Describe fully the single transformation that maps shape B onto shape C.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 reflection✓ 1
B1 in the x-axis (or the line y = 0)✓ 1
Part (b)
B1 rotation✓ 1
B1 180°✓ 1
B1 centre (0, 0)✓ 1

Exam tip: Give only one transformation when the question says 'single'. A rotation needs three things: the word rotation, the angle (and direction, unless it is 180°), and the centre.

5IGME-062-00-01Vectors and vector geometry Paper FEasy3 marks

The vectors a and b are given by
a = \(\begin{pmatrix}3\\-2\end{pmatrix}\) and b = \(\begin{pmatrix}-1\\5\end{pmatrix}\)

(a)

Find a + b as a column vector.

(1)
(b)

Find 3a − 2b as a column vector.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1: \(\begin{pmatrix}2\\3\end{pmatrix}\)✓ 1
Part (b)
M1: 3a = \(\begin{pmatrix}9\\-6\end{pmatrix}\) or 2b = \(\begin{pmatrix}-2\\10\end{pmatrix}\)✓ 1
A1: \(\begin{pmatrix}11\\-16\end{pmatrix}\)✓ 1

Common error: Subtracting a negative component: 9 − (−2) = 11, not 7. Write each component calculation out separately.

6IGME-062-00-02Vectors and vector geometry Paper FEasy3 marks

The point A has coordinates (2, 7) and the point B has coordinates (6, −1).

(a)

Write \(\overrightarrow{AB}\) as a column vector.

(2)
(b)

The point C is such that \(\overrightarrow{AC}\) = \(\begin{pmatrix}-3\\4\end{pmatrix}\). Find the coordinates of C.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1: subtracts coordinates of A from B, e.g. 6 − 2 or −1 − 7✓ 1
A1: \(\begin{pmatrix}4\\-8\end{pmatrix}\)✓ 1
Part (b)
B1: (−1, 11)✓ 1

Common error: \(\overrightarrow{AB}\) means 'from A to B', so it is B − A. Working out A − B gives the vector in the opposite direction.

7IGME-062-00-04Vectors and vector geometry Paper FEasy3 marks

\(\begin{pmatrix}k\\3\end{pmatrix}\) + 2\(\begin{pmatrix}4\\m\end{pmatrix}\) = \(\begin{pmatrix}13\\-5\end{pmatrix}\)

(a)

Find the value of k and the value of m.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1: forms an equation for either component, k + 8 = 13 or 3 + 2m = −5✓ 1
A1: k = 5✓ 1
A1: m = −4✓ 1

Exam tip: Treat the top and bottom rows as two separate equations. The scalar 2 multiplies both components of its vector.

8IGME-062-00-05Vectors and vector geometry Paper FEasy3 marks

Triangle T has vertices P(1, 2), Q(4, 2) and R(1, 6).
Triangle T is translated by the vector \(\begin{pmatrix}-5\\3\end{pmatrix}\) to give triangle U.

(a)

Find the coordinates of the vertices of triangle U.

(2)
(b)

Write down the column vector of the translation that maps triangle U back onto triangle T.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1: at least one vertex correct, e.g. (−4, 5)✓ 1
A1: all three correct: (−4, 5), (−1, 5), (−4, 9)✓ 1
Part (b)
B1: \(\begin{pmatrix}5\\-3\end{pmatrix}\)✓ 1

Exam tip: The top number moves points horizontally (negative = left), the bottom number vertically (positive = up). The inverse translation reverses both signs.

9IGME-062-01-06Transformations described in words Paper FMedium4 marks

Triangle P has vertices at (1, 1), (3, 1) and (1, 2).
Triangle Q has vertices at (1, −1), (1, 1) and (0, −1).
Triangle P is mapped onto triangle Q by a single rotation.

(a)

Describe fully this rotation.

(3)
(b)

Write down the coordinates of the only point that is not moved by this rotation.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 rotation 90°✓ 1
B1 anticlockwise✓ 1
B1 centre (2, 0)✓ 1
Part (b)
B1 (2, 0)✓ 1

Exam tip: To find the centre, take one vertex and its image. The centre is the same distance from both, and the lines joining the centre to each one meet at the angle of rotation. Check your answer with a second vertex.

10IGME-062-01-07Transformations described in words Paper FMedium5 marks

Triangle R has vertices at (5, 7), (9, 7) and (9, 11).
Triangle R is enlarged with scale factor \(\frac{1}{2}\) and centre (1, 3) to give triangle S.

(a)

Find the coordinates of the vertices of triangle S.

(3)
(b)

Find the area of triangle S.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 for finding the vector from (1, 3) to a vertex and halving it, e.g. (5, 7) → \(\begin{pmatrix} 4 \\ 4 \end{pmatrix}\) → \(\begin{pmatrix} 2 \\ 2 \end{pmatrix}\)✓ 1
A1 for two correct vertices from (3, 5), (5, 5), (5, 7)✓ 1
A1 for all three correct✓ 1
Part (b)
M1 for ½ × 2 × 2, or (½ × 4 × 4) × (½)2✓ 1
A1 2 square units✓ 1

Common error: Do not halve the coordinates of each vertex directly. That would be an enlargement about the origin. Halve the distances measured from the centre (1, 3).

11IGME-062-01-08Transformations described in words Paper FMedium5 marks

Triangle A has vertices at (2, 1), (4, 1) and (4, 2).
Triangle B has vertices at (4, 3), (10, 3) and (10, 6).
Triangle B is the image of triangle A under a single transformation.

(a)

Describe fully the single transformation that maps triangle A onto triangle B.

(3)
(b)

The perimeter of triangle A is \(3+\sqrt{5}\) units.
Find the perimeter of triangle B. Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 enlargement✓ 1
B1 scale factor 3✓ 1
B1 centre (1, 0)✓ 1
Part (b)
M1 for 3 × (3 + √5)✓ 1
A1 15.7 units✓ 1

Exam tip: To find the centre of enlargement, draw lines through matching vertices of the two shapes and see where they cross. You can also work it out with straight-line equations. Here the line through (2, 1) and (4, 3) meets the line through (4, 1) and (10, 3) at (1, 0).

12IGME-062-01-10Transformations described in words Paper FMedium7 marks

Triangle E has vertices at (1, 2), (3, 2) and (3, 3).
Triangle E is reflected in the line y = x to give triangle F.
Triangle F is then reflected in the x-axis to give triangle G.

(a)

Write down the coordinates of the vertices of triangle F.

(2)
(b)

Write down the coordinates of the vertices of triangle G.

(2)
(c)

Describe fully the single transformation that maps triangle E onto triangle G.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
B2 for (2, 1), (2, 3), (3, 3)✓ 2
(B1 for two correct vertices)—
Part (b)
B2 for (2, −1), (2, −3), (3, −3)✓ 2
(B1 ft from their F for two correct vertices)—
Part (c)
B1 rotation✓ 1
B1 90° clockwise✓ 1
B1 centre (0, 0)✓ 1

Exam tip: Two reflections in lines that cross always combine to give a rotation about the point where the lines meet. Here the two mirror lines meet at the origin.

13IGME-062-01-11Transformations described in words Paper FMedium4 marks

A translation maps the point A(−3, 5) onto the point A′(4, −1).

(a)

Write down the column vector for this translation.

(1)
(b)

The same translation maps the point B(2, 2) onto the point B′.
Find the coordinates of B′.

(1)
(c)

Find the distance that each point moves under this translation. Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 \(\begin{pmatrix} 7 \\ -6 \end{pmatrix}\)✓ 1
Part (b)
B1 (9, −4)✓ 1
Part (c)
M1 for √(72 + 62) or √85✓ 1
A1 9.22 units✓ 1

Common error: Subtract in the right order: the vector is (image − original). Working out (−3 − 4) instead of (4 − (−3)) gives a vector that points the wrong way.

14IGME-062-00-08Vectors and vector geometry Paper FMedium4 marks

a = \(\begin{pmatrix}2\\-3\end{pmatrix}\), b = \(\begin{pmatrix}x\\12\end{pmatrix}\) and c = \(\begin{pmatrix}6\\y\end{pmatrix}\)

(a)

b is parallel to a. Find the value of x.

(2)
(b)

a + c is parallel to \(\begin{pmatrix}1\\1\end{pmatrix}\). Find the value of y.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1: recognises b = −4a (12 ÷ −3 = −4)✓ 1
A1: x = −8✓ 1
Part (b)
M1: a + c = \(\begin{pmatrix}8\\y-3\end{pmatrix}\) and components set equal, y − 3 = 8✓ 1
A1: y = 11✓ 1

Exam tip: Parallel vectors are scalar multiples of each other. Find the multiplier from the component you know, then apply it to the other component.

15IGME-062-01-12Transformations described in words Paper FHard6 marks

Triangle M has vertices at (1, 1), (3, 1) and (1, 2).
Triangle N has vertices at (3, −1), (3, 1) and (4, −1).
Triangle M is mapped onto triangle N by a reflection. The vertex (1, 1) maps to (3, −1), and the vertex (3, 1) maps to itself.

(a)

Find the equation of the mirror line.

(3)
(b)

Explain why the vertex (3, 1) is not moved by the reflection.

(1)
(c)

The point (−4, 7) is reflected in the same mirror line.
Find the coordinates of its image.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 for the midpoint of (1, 1) and (3, −1), which is (2, 0)✓ 1
M1 for using two points on the line, e.g. (2, 0) and (3, 1), giving gradient 1✓ 1
A1 y = x − 2✓ 1
Part (b)
B1 because (3, 1) lies on the mirror line (1 = 3 − 2)✓ 1
Part (c)
M1 for a method, e.g. the image is (y + 2, x − 2), or using the perpendicular through (−4, 7) with gradient −1 and its midpoint on the line✓ 1
A1 (9, −6)✓ 1

Exam tip: The mirror line is the perpendicular bisector of the segment joining any point to its image. Check your answer: the midpoint of (−4, 7) and (9, −6) is (2.5, 0.5), and this point lies on y = x − 2.

16IGME-062-00-03Vectors and vector geometry Paper HEasy4 marks

p = \(\begin{pmatrix}5\\-12\end{pmatrix}\) and q = \(\begin{pmatrix}1\\4\end{pmatrix}\)

(a)

Find |p|.

(2)
(b)

Find |p + q|.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1: \(\sqrt{5^2 + (-12)^2}\)✓ 1
A1: 13✓ 1
Part (b)
M1: p + q = \(\begin{pmatrix}6\\-8\end{pmatrix}\) and Pythagoras used, \(\sqrt{6^2+8^2}\)✓ 1
A1: 10✓ 1

Common error: |p + q| is not |p| + |q|. Add the vectors first, then find the magnitude.

17IGME-062-01-09Transformations described in words Paper HMedium7 marks

Triangle J has vertices at (2, 1), (4, 1) and (2, 4).
Triangle J is enlarged with scale factor −2 and centre (1, 0) to give triangle K.

(a)

Find the coordinates of the vertices of triangle K.

(3)
(b)

Find the area of triangle K.

(2)
(c)

Describe fully the single transformation that maps triangle K onto triangle J.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 for multiplying the vector from (1, 0) to a vertex by −2, e.g. \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\) → \(\begin{pmatrix} -2 \\ -2 \end{pmatrix}\)✓ 1
A1 for two correct vertices from (−1, −2), (−5, −2), (−1, −8)✓ 1
A1 for all three correct✓ 1
Part (b)
M1 for (½ × 2 × 3) × (−2)2, or ½ × 4 × 6✓ 1
A1 12 square units✓ 1
Part (c)
B1 enlargement, centre (1, 0)✓ 1
B1 scale factor \(-\frac{1}{2}\)✓ 1

Common error: A negative scale factor puts the image on the opposite side of the centre. The area still scales by the square of the scale factor, which is positive: (−2)² = 4.

18IGME-062-00-06Vectors and vector geometry Paper HMedium4 marks

O, A and B are three points that do not lie on a straight line.
\(\overrightarrow{OA}\) = a and \(\overrightarrow{OB}\) = b.
M is the midpoint of AB.
N is the point on OB such that ON : NB = 1 : 2.

(a)

Find \(\overrightarrow{OM}\) in terms of a and b. Give your answer in its simplest form.

(2)
(b)

Find \(\overrightarrow{MN}\) in terms of a and b. Give your answer in its simplest form.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1: \(\overrightarrow{OM}\) = a + ½(b − a) or equivalent route✓ 1
A1: ½a + ½b✓ 1
Part (b)
M1: \(\overrightarrow{ON}\) = ⅓b and \(\overrightarrow{MN}\) = \(\overrightarrow{ON}\) − \(\overrightarrow{OM}\)✓ 1
A1: −½a − ⅙b✓ 1

Common error: A ratio of 1 : 2 means ON is \(\frac{1}{3}\) of OB, not \(\frac{1}{2}\). Add the parts of the ratio to find the fraction.

19IGME-062-00-09Vectors and vector geometry Paper HMedium4 marks

OABC is a parallelogram with vertices in that order.
\(\overrightarrow{OA}\) = a and \(\overrightarrow{OC}\) = c.

(a)

Find \(\overrightarrow{AC}\) in terms of a and c.

(1)
(b)

Find \(\overrightarrow{OB}\) in terms of a and c.

(1)
(c)

E is the point on OB such that OE : EB = 2 : 1.
Find \(\overrightarrow{AE}\) in terms of a and c. Give your answer in its simplest form.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1: c − a✓ 1
Part (b)
B1: a + c✓ 1
Part (c)
M1: \(\overrightarrow{AE}\) = −a + ⅔(a + c)✓ 1
A1: −⅓a + ⅔c✓ 1

Exam tip: In a parallelogram, opposite sides are equal vectors: \(\overrightarrow{AB}\) = \(\overrightarrow{OC}\) = c. Use this to build any route through the shape.

20IGME-062-00-10Vectors and vector geometry Paper HMedium5 marks

A drone starts at point O. Distances are in metres.
It flies with displacement \(\begin{pmatrix}12\\5\end{pmatrix}\) and then with displacement \(\begin{pmatrix}-4\\10\end{pmatrix}\).

(a)

Find the resultant displacement of the drone from O as a column vector.

(1)
(b)

Find the distance of the drone from O.

(2)
(c)

The drone then flies in a straight line directly back to O at a constant speed of 2.5 m/s.
Find the time it takes.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1: \(\begin{pmatrix}8\\15\end{pmatrix}\)✓ 1
Part (b)
M1: \(\sqrt{8^2+15^2}\)✓ 1
A1: 17 m✓ 1
Part (c)
M1: 17 ÷ 2.5✓ 1
A1: 6.8 s✓ 1

Common error: The distance back is the magnitude of the resultant, 17 m, not the total distance flown along the two legs.

21IGME-062-00-11Vectors and vector geometry Paper HMedium4 marks

A is the point (1, 3) and B is the point (7, −5).
M is the point on AB such that AM : MB = 1 : 3.

(a)

Use vectors to find the coordinates of M.

(2)
(b)

Find the length of AM.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1: \(\overrightarrow{AB}\) = \(\begin{pmatrix}6\\-8\end{pmatrix}\) and \(\overrightarrow{AM}\) = ¼\(\overrightarrow{AB}\) = \(\begin{pmatrix}1.5\\-2\end{pmatrix}\)✓ 1
A1: M = (2.5, 1)✓ 1
Part (b)
M1: \(\sqrt{1.5^2+2^2}\)✓ 1
A1: 2.5✓ 1

Common error: AM : MB = 1 : 3 means AM is \(\frac{1}{4}\) of AB, not \(\frac{1}{3}\). Remember to add \(\overrightarrow{AM}\) to the position of A to get the coordinates of M.

22IGME-062-00-07Vectors and vector geometry Paper HMedium6 marks

v = \(\begin{pmatrix}-6\\8\end{pmatrix}\)

(a)

Find |v|.

(2)
(b)

w = kv, where k is a positive number, and |w| = 25.
Find w as a column vector.

(2)
(c)

Find the angle, measured anticlockwise from the positive x-direction, that v makes. Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1: \(\sqrt{(-6)^2+8^2}\)✓ 1
A1: 10✓ 1
Part (b)
M1: k = 25 ÷ 10 = 2.5✓ 1
A1: \(\begin{pmatrix}-15\\20\end{pmatrix}\)✓ 1
Part (c)
M1: a correct method, e.g. 180 − tan−1(8/6), or cos−1(−6/10), or 90 + tan−1(6/8)✓ 1
A1: 127°✓ 1

Exam tip: A vector with a negative x-component and a positive y-component points into the second quadrant, so the angle must lie between 90° and 180°.

23IGME-062-00-12Vectors and vector geometry Paper HHard7 marks

OAB is a triangle.
\(\overrightarrow{OA}\) = a and \(\overrightarrow{OB}\) = b.
P is the point on OA such that OP : PA = 2 : 1.
Q is the midpoint of AB.
R is the point on OB extended such that B is the midpoint of OR.

(a)

Find \(\overrightarrow{PQ}\) in terms of a and b. Give your answer in its simplest form.

(3)
(b)

Use vectors to show that P, Q and R lie on a straight line.

(3)
(c)

Write down the ratio PQ : QR.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1: \(\overrightarrow{OP}\) = ⅔a or \(\overrightarrow{OQ}\) = ½a + ½b✓ 1
M1: \(\overrightarrow{PQ}\) = \(\overrightarrow{OQ}\) − \(\overrightarrow{OP}\) or equivalent route✓ 1
A1: −⅙a + ½b✓ 1
Part (b)
M1: \(\overrightarrow{OR}\) = 2b✓ 1
M1: \(\overrightarrow{PR}\) = −⅔a + 2b✓ 1
A1: \(\overrightarrow{PR}\) = 4\(\overrightarrow{PQ}\), so PR is parallel to PQ and they share point P, hence P, Q, R collinear✓ 1
Part (c)
B1: 1 : 3✓ 1

Exam tip: To prove collinearity, show one vector is a multiple of the other AND state that they share a common point. Parallel alone is not enough.

24IGME-062-00-13Vectors and vector geometry Paper HHard7 marks

OAB is a triangle.
\(\overrightarrow{OA}\) = a and \(\overrightarrow{OB}\) = b.
C is the midpoint of OB.
D is the point on AB such that AD : DB = 2 : 1.
The lines OD and AC intersect at the point E.

(a)

Find \(\overrightarrow{OD}\) in terms of a and b. Give your answer in its simplest form.

(2)
(b)

\(\overrightarrow{OE}\) = μ\(\overrightarrow{OD}\) and \(\overrightarrow{AE}\) = t\(\overrightarrow{AC}\), where μ and t are scalars.
Write two different expressions for \(\overrightarrow{OE}\) in terms of a, b, μ and t.

(2)
(c)

Hence find the ratio OE : ED.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1: a + ⅔(b − a)✓ 1
A1: ⅓a + ⅔b✓ 1
Part (b)
B1: \(\overrightarrow{OE}\) = ⅓μa + ⅔μb✓ 1
B1: \(\overrightarrow{OE}\) = a + t(½b − a) = (1 − t)a + ½tb✓ 1
Part (c)
M1: equates coefficients: ⅓μ = 1 − t and ⅔μ = ½t✓ 1
M1: solves to get μ = 3/5 (with t = 4/5)✓ 1
A1: OE : ED = 3 : 2✓ 1

Exam tip: Since a and b are not parallel, two expressions for the same vector must have equal coefficients of a and equal coefficients of b. This gives simultaneous equations.

25IGME-062-00-14Vectors and vector geometry Paper HHard6 marks

ABCDEF is a regular hexagon with vertices in that order.
\(\overrightarrow{AB}\) = p and \(\overrightarrow{BC}\) = q.

(a)

Find \(\overrightarrow{CD}\) in terms of p and q.

(1)
(b)

Find \(\overrightarrow{AD}\) in terms of p and q. Give your answer in its simplest form.

(2)
(c)

M is the midpoint of CD.
Find \(\overrightarrow{AM}\) in terms of p and q. Give your answer in its simplest form.

(2)
(d)

Each side of the hexagon is 5 cm long. Write down the length of AD.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1: q − p✓ 1
Part (b)
M1: p + q + (q − p)✓ 1
A1: 2q✓ 1
Part (c)
M1: p + q + ½(q − p)✓ 1
A1: ½p + \(\frac{3}{2}\)q✓ 1
Part (d)
B1: 10 cm✓ 1

Exam tip: In a regular hexagon, the long diagonal AD is parallel to BC and twice its length. Use this to check that \(\overrightarrow{AD}\) = 2q.

26IGME-062-01-14Transformations described in words Paper HHard6 marks

A shape has area 12 square units.
It is enlarged with a positive scale factor k and centre C(2, 3). The image has area 75 square units.
One vertex of the original shape is P(4, 4), and its image is P′.

(a)

Find the value of k.

(2)
(b)

Find the coordinates of P′.

(2)
(c)

Find the distance PP′. Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 for k2 = 75 ÷ 12 = 6.25✓ 1
A1 k = 2.5✓ 1
Part (b)
M1 for \(\overrightarrow{CP} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}\), multiplied by 2.5 and added to (2, 3)✓ 1
A1 (7, 5.5)✓ 1
Part (c)
M1 for √(32 + 1.52) or √11.25, or for (2.5 − 1) × √(22 + 12) = 1.5√5✓ 1
A1 3.35 units✓ 1

Common error: 75 ÷ 12 = 6.25 is the area scale factor, not the length scale factor. Take its square root to get k = 2.5. Do not use 6.25 as k.

© ExaminerPrep · examinerprep.com · Original material by Evan Kotronis. ExaminerPrep is an independent tutoring service. It has been developed independently from and is not endorsed by the International Baccalaureate Organization. "International Baccalaureate", "IB" and "IB Diploma Programme" are registered trademarks of the IBO, used here for descriptive purposes only. Pearson Edexcel and Cambridge are trademarks of their owners; ExaminerPrep is not affiliated with any exam board.