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Cambridge IGCSE Mathematics 0580 · Core and Extended · Topic 7

Transformations and vectors

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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 →

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+ Add to test✓ In your test1IGMC-047-00-01Reflections, rotations, translations and enlargements described in words Paper CEasy4 marks

No calculator

Triangle A has vertices at (1, 1), (3, 1) and (1, 4).

(a)

Triangle A is reflected in the x-axis to give triangle B. Write down the coordinates of the vertices of triangle B.

(2)

(b)

Triangle A is translated by the vector \(\begin{pmatrix}-4\\2\end{pmatrix}\) to give triangle C. Write down the coordinates of the vertices of triangle C.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(1, −1), (3, −1), (1, −4)	—
B1 for at least two vertices correct	✓ 1
B1 for all three vertices correct	✓ 1
Part (b)
(−3, 3), (−1, 3), (−3, 6)	—
B1 for at least two vertices correct	✓ 1
B1 for all three vertices correct	✓ 1

Common error: Reflecting in the x-axis changes the sign of the y-coordinate, not the x-coordinate. The line stays fixed, so points move up or down across it.

+ Add to test✓ In your test2IGMC-047-00-02Reflections, rotations, translations and enlargements described in words Paper CEasy3 marks

No calculator

The point P has coordinates (2, 5).

(a)

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

(1)

(b)

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

(1)

(c)

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

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for (5, −2)	✓ 1
Part (b)
B1 for (−2, −5)	✓ 1
Part (c)
B1 for (5, 2)	✓ 1

Exam tip: Check a clockwise rotation quickly: a point in the first quadrant rotated 90° clockwise about the origin must land in the fourth quadrant, with positive x and negative y.

+ Add to test✓ In your test3IGMC-047-00-03Reflections, rotations, translations and enlargements described in words Paper CEasy4 marks

No calculator

Triangle A has vertices (2, 1), (4, 1) and (2, 2).
Triangle B has vertices (2, −1), (4, −1) and (2, −2).
Triangle C has vertices (−3, 4), (−1, 4) and (−3, 5).

(a)

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

(2)

(b)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for reflection	✓ 1
B1 for in the x-axis (or y = 0)	✓ 1
Part (b)
B1 for translation	✓ 1
B1 for vector \(\begin{pmatrix}-5\\3\end{pmatrix}\)	✓ 1

Exam tip: A full description of a reflection needs the equation of the mirror line, and a translation needs a column vector. Writing 'move 5 left and 3 up' instead of a column vector risks losing the mark.

+ Add to test✓ In your test4IGMC-047-00-04Reflections, rotations, translations and enlargements described in words Paper CEasy4 marks

No calculator

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

(a)

Write down the coordinates of the vertices of triangle U.

(2)

(b)

Find the area of triangle U.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(3, 6), (6, 6), (3, 9)	—
B1 for at least two vertices correct	✓ 1
B1 for all three vertices correct	✓ 1
Part (b)
M1 for area of T = ½ × 1 × 1 = 0.5 and multiplying by 3², or for ½ × 3 × 3 using triangle U	✓ 1
A1 for 4.5 square units	✓ 1

Common error: Lengths are multiplied by the scale factor, but areas are multiplied by the square of the scale factor. The area of U is 9 times the area of T, not 3 times.

+ Add to test✓ In your test5IGMC-047-00-10Reflections, rotations, translations and enlargements described in words Paper CEasy5 marks

No calculator

Triangle A has vertices (1, 1), (3, 1) and (1, 2).
Triangle B has vertices (−1, 1), (−1, 3) and (−2, 1).

(a)

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

(3)

(b)

Triangle A is reflected in the line y = −x. Write down the coordinates of the vertices of the image.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for rotation	✓ 1
B1 for 90° anticlockwise (accept 270° clockwise)	✓ 1
B1 for centre (0, 0) or origin	✓ 1
Part (b)
(−1, −1), (−1, −3), (−2, −1)	—
B1 for at least two vertices correct	✓ 1
B1 for all three vertices correct	✓ 1

Common error: 'Rotation of 90° about the origin' is not a full description. The direction (clockwise or anticlockwise) is needed for any rotation that is not 180°.

+ Add to test✓ In your test6IGMC-047-00-05Reflections, rotations, translations and enlargements described in words Paper CMedium7 marks

No calculator

Quadrilateral Q has vertices (3, 1), (5, 1), (5, 3) and (4, 3).

(a)

Q is reflected in the line x = 2 to give quadrilateral R. Write down the coordinates of the vertices of R.

(2)

(b)

R is reflected in the line y = −1 to give quadrilateral S. Write down the coordinates of the vertices of S.

(2)

(c)

Describe fully the single transformation that maps Q onto S.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(1, 1), (−1, 1), (−1, 3), (0, 3)	—
B1 for at least two vertices correct	✓ 1
B1 for all four vertices correct	✓ 1
Part (b)
(1, −3), (−1, −3), (−1, −5), (0, −5)	—
B1 for at least two vertices correct, follow through from their R	✓ 1
B1 for all four vertices correct	✓ 1
Part (c)
B1 for rotation	✓ 1
B1 for 180°	✓ 1
B1 for centre (2, −1)	✓ 1

Exam tip: Reflections in two perpendicular lines combine to give a 180° rotation about the point where the lines cross. Here x = 2 and y = −1 meet at (2, −1). For a 180° rotation no direction is needed.

+ Add to test✓ In your test7IGMC-047-00-06Reflections, rotations, translations and enlargements described in words Paper CMedium6 marks

No calculator

Triangle T has vertices (1, 2), (3, 2) and (3, 3).

(a)

Triangle T is rotated 90° anticlockwise about the point (1, 0) to give triangle V. Find the coordinates of the vertices of triangle V.

(3)

(b)

Describe fully the single transformation that maps triangle V back onto triangle T.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Relative to (1, 0), the vertices of T are (0, 2), (2, 2) and (2, 3). A rotation of 90° anticlockwise sends (x, y) to (−y, x), giving (−2, 0), (−2, 2) and (−3, 2). Adding (1, 0) back gives the vertices of V.	—
B1 for (−1, 0)	✓ 1
B1 for (−1, 2)	✓ 1
B1 for (−2, 2)	✓ 1
Part (b)
B1 for rotation	✓ 1
B1 for 90° clockwise (accept 270° anticlockwise)	✓ 1
B1 for centre (1, 0)	✓ 1

Common error: Rotating about the origin when the centre is (1, 0) gives the wrong images. Work out each vertex's displacement from the centre, rotate that displacement, then add it back on to the centre.

+ Add to test✓ In your test8IGMC-047-00-07Reflections, rotations, translations and enlargements described in words Paper CMedium5 marks

No calculator

Triangle P has vertices (2, 1), (4, 1) and (2, 4).
Triangle Q has vertices (−2, −1), (4, −1) and (−2, 8).

(a)

Describe fully the single transformation that maps triangle P onto triangle Q.

(3)

(b)

Find the area of triangle Q.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for enlargement	✓ 1
B1 for scale factor 3	✓ 1
B1 for centre (4, 2)	✓ 1
Part (b)
M1 for ½ × 6 × 9, or for area of P = 3 multiplied by 3²	✓ 1
A1 for 27 square units	✓ 1

Exam tip: To find the centre of enlargement, draw or imagine lines through matching vertices, such as (2, 1) to (−2, −1) and (4, 1) to (4, −1). The centre is where these lines meet. Check that the image distances from the centre are 3 times the original distances.

+ Add to test✓ In your test9IGMC-047-00-09Reflections, rotations, translations and enlargements described in words Paper CMedium6 marks

Calculator

A shape is translated by the vector \(\begin{pmatrix}3\\-2\end{pmatrix}\) and then by the vector \(\begin{pmatrix}-5\\4\end{pmatrix}\).

(a)

Write down the vector of the single translation that has the same effect as these two translations.

(1)

(b)

After both translations, one vertex of the shape is at (1, 7). Find the coordinates of this vertex before the translations.

(2)

(c)

Write down the vector of the translation that maps the final shape back onto the original shape.

(1)

(d)

(Extended) Find the distance moved by each point under the single translation that has the same effect as both translations. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for \(\begin{pmatrix}-2\\2\end{pmatrix}\)	✓ 1
Part (b)
M1 for (1 − (−2), 7 − 2) or reversing each translation in turn	✓ 1
A1 for (3, 5)	✓ 1
Part (c)
B1 for \(\begin{pmatrix}2\\-2\end{pmatrix}\), follow through from their single translation vector	✓ 1
Part (d)
M1 for \(\sqrt{(-2)^2 + 2^2}\)	✓ 1
A1 for 2.83 units (accept \(2\sqrt{2}\) or \(\sqrt{8}\))	✓ 1

Exam tip: Add translation vectors component by component: top numbers together, bottom numbers together. To undo a translation, change the sign of both components.

+ Add to test✓ In your test10IGMC-047-00-12Reflections, rotations, translations and enlargements described in words Paper CMedium6 marks

No calculator

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

(a)

T is rotated 90° clockwise about the origin to give triangle U. Write down the coordinates of the vertices of U.

(2)

(b)

U is reflected in the x-axis to give triangle V. Write down the coordinates of the vertices of V.

(2)

(c)

Describe fully the single transformation that maps triangle T onto triangle V.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(2, −1), (2, −4), (3, −4)	—
B1 for at least two vertices correct	✓ 1
B1 for all three vertices correct	✓ 1
Part (b)
(2, 1), (2, 4), (3, 4)	—
B1 for at least two vertices correct, follow through from their U	✓ 1
B1 for all three vertices correct	✓ 1
Part (c)
B1 for reflection	✓ 1
B1 for in the line y = x	✓ 1

Common error: Two transformations combined must be described as one transformation. Answers such as 'rotation then reflection' score nothing when a single transformation is asked for. Compare the coordinates of T and V directly.

+ Add to test✓ In your test11IGMC-047-01-01Vectors, magnitude and vector geometry Paper EEasy4 marks

No calculator

\(\mathbf{a}=\begin{pmatrix}3\\-4\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}-1\\2\end{pmatrix}\).

(a)

Find \(2\mathbf{a}-3\mathbf{b}\) as a column vector.

(2)

(b)

Find \(|\mathbf{a}|\).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: \(2\mathbf{a}=\begin{pmatrix}6\\-8\end{pmatrix}\) and \(3\mathbf{b}=\begin{pmatrix}-3\\6\end{pmatrix}\) seen, or one component correct	✓ 1
A1: \(\begin{pmatrix}9\\-14\end{pmatrix}\)	✓ 1
Part (b)
M1: \(\sqrt{3^2+(-4)^2}\)	✓ 1
A1: 5	✓ 1

Common error: Subtracting a negative component: \(-3\times(-1)=+3\), so the top entry is \(6+3=9\), not \(6-3\).

+ Add to test✓ In your test12IGMC-047-01-02Vectors, magnitude and vector geometry Paper EEasy4 marks

Calculator

\(\mathbf{p}=\begin{pmatrix}7\\-2\end{pmatrix}\) and \(\mathbf{p}+\mathbf{q}=\begin{pmatrix}2\\5\end{pmatrix}\).

(a)

Find \(\mathbf{q}\) as a column vector.

(2)

(b)

Find \(|\mathbf{p}|\), giving your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: \(\mathbf{q}=\begin{pmatrix}2\\5\end{pmatrix}-\begin{pmatrix}7\\-2\end{pmatrix}\)	✓ 1
A1: \(\begin{pmatrix}-5\\7\end{pmatrix}\)	✓ 1
Part (b)
M1: \(\sqrt{7^2+(-2)^2}=\sqrt{53}\)	✓ 1
A1: 7.28	✓ 1

Exam tip: Squaring a negative component always gives a positive value, so the magnitude is never smaller because of a minus sign.

+ Add to test✓ In your test13IGMC-047-01-03Vectors, magnitude and vector geometry Paper EEasy4 marks

No calculator

A is the point (2, 5) and B is the point (−4, 1).

(a)

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

(1)

(b)

\(\overrightarrow{AC}=\begin{pmatrix}3\\-2\end{pmatrix}\). Find the coordinates of C.

(1)

(c)

Find \(|\overrightarrow{AB}|\), giving your answer in the form \(k\sqrt{13}\).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: \(\begin{pmatrix}-6\\-4\end{pmatrix}\)	✓ 1
Part (b)
B1: (5, 3)	✓ 1
Part (c)
M1: \(\sqrt{(-6)^2+(-4)^2}=\sqrt{52}\)	✓ 1
A1: \(2\sqrt{13}\)	✓ 1

Common error: \(\overrightarrow{AB}\) is B minus A, not A minus B. Reversing the order gives the right length but the wrong direction.

+ Add to test✓ In your test14IGMC-047-01-04Vectors, magnitude and vector geometry Paper EEasy3 marks

No calculator

\(\mathbf{a}=\begin{pmatrix}6\\-9\end{pmatrix}\).

(a)

Show that \(\mathbf{c}=\begin{pmatrix}-4\\6\end{pmatrix}\) is parallel to \(\mathbf{a}\).

(1)

(b)

\(\mathbf{b}=\begin{pmatrix}k\\3\end{pmatrix}\) is parallel to \(\mathbf{a}\). Find the value of \(k\).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: \(\mathbf{c}=-\frac{2}{3}\mathbf{a}\) (or \(\mathbf{a}=-\frac{3}{2}\mathbf{c}\)) stated, so c is a multiple of a	✓ 1
Part (b)
M1: \(\mathbf{b}=-\frac{1}{3}\mathbf{a}\) or \(\frac{k}{6}=\frac{3}{-9}\)	✓ 1
A1: \(k=-2\)	✓ 1

Exam tip: Two vectors are parallel when one is a scalar multiple of the other; the scalar may be negative or a fraction.

+ Add to test✓ In your test15IGMC-047-01-05Vectors, magnitude and vector geometry Paper EEasy4 marks

No calculator

Answer both parts without a calculator.

(a)

Find the magnitude of \(\begin{pmatrix}-8\\15\end{pmatrix}\).

(2)

(b)

The vector \(\begin{pmatrix}p\\12\end{pmatrix}\) has magnitude 13 and \(p>0\). Find \(p\).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: \(\sqrt{64+225}=\sqrt{289}\)	✓ 1
A1: 17	✓ 1
Part (b)
M1: \(p^2+144=169\)	✓ 1
A1: \(p=5\)	✓ 1

Common error: Writing \(p+12=13\) is wrong: magnitudes combine through squares, so form \(p^2+12^2=13^2\).

+ Add to test✓ In your test16IGMC-047-00-08Reflections, rotations, translations and enlargements described in words Paper EMedium6 marks

No calculator

Triangle A has vertices (2, 2), (4, 2) and (4, 6).
Triangle A is enlarged with scale factor \(-\frac{1}{2}\) and centre (0, 0) to give triangle B.

(a)

Find the coordinates of the vertices of triangle B.

(2)

(b)

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

(2)

(c)

The area of triangle A is 4 square units. Find the area of triangle B.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(−1, −1), (−2, −1), (−2, −3)	—
B1 for at least two vertices correct	✓ 1
B1 for all three vertices correct	✓ 1
Part (b)
B1 for enlargement with centre (0, 0)	✓ 1
B1 for scale factor −2	✓ 1
Part (c)
M1 for 4 × \(\left(\frac{1}{2}\right)^2\) or \(\frac{1}{2}\) × 1 × 2 using the vertices of B	✓ 1
A1 for 1 square unit	✓ 1

Common error: A negative scale factor does not give a negative area. The area factor is the square of the scale factor: \((-\frac{1}{2})^2 = \frac{1}{4}\).

+ Add to test✓ In your test17IGMC-047-01-06Vectors, magnitude and vector geometry Paper EMedium6 marks

No calculator

OABC is a parallelogram with vertices in that order. \(\overrightarrow{OA}=\mathbf{a}\) and \(\overrightarrow{OC}=\mathbf{c}\). M is the midpoint of AB.
Give each answer in terms of \(\mathbf{a}\) and \(\mathbf{c}\), in its simplest form.

(a)

Find \(\overrightarrow{OB}\).

(1)

(b)

Find \(\overrightarrow{AC}\).

(1)

(c)

Find \(\overrightarrow{OM}\).

(2)

(d)

Find \(\overrightarrow{CM}\).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: \(\mathbf{a}+\mathbf{c}\)	✓ 1
Part (b)
B1: \(\mathbf{c}-\mathbf{a}\)	✓ 1
Part (c)
M1: any correct route, e.g. \(\overrightarrow{OA}+\frac{1}{2}\overrightarrow{AB}\) with \(\overrightarrow{AB}=\mathbf{c}\), or \(\overrightarrow{OB}-\frac{1}{2}\mathbf{c}\)	✓ 1
A1: \(\mathbf{a}+\frac{1}{2}\mathbf{c}\)	✓ 1
Part (d)
M1: any correct route, e.g. \(\overrightarrow{CO}+\overrightarrow{OM}=-\mathbf{c}+\mathbf{a}+\frac{1}{2}\mathbf{c}\), or \(\overrightarrow{CB}+\overrightarrow{BM}=\mathbf{a}-\frac{1}{2}\mathbf{c}\)	✓ 1
A1: \(\mathbf{a}-\frac{1}{2}\mathbf{c}\)	✓ 1

Exam tip: In a parallelogram, opposite sides are equal vectors: \(\overrightarrow{AB}=\overrightarrow{OC}=\mathbf{c}\) and \(\overrightarrow{CB}=\overrightarrow{OA}=\mathbf{a}\). Use this to move around the shape.

+ Add to test✓ In your test18IGMC-047-01-07Vectors, magnitude and vector geometry Paper EMedium5 marks

No calculator

In triangle OAB, \(\overrightarrow{OA}=\mathbf{a}\) and \(\overrightarrow{OB}=\mathbf{b}\). P lies on AB such that AP : PB = 1 : 3. Q lies on OB such that OQ : QB = 1 : 3.

(a)

Find \(\overrightarrow{AB}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).

(1)

(b)

Find \(\overrightarrow{OP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\), in its simplest form.

(2)

(c)

Show that QP is parallel to OA.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: \(\mathbf{b}-\mathbf{a}\)	✓ 1
Part (b)
M1: \(\mathbf{a}+\frac{1}{4}(\mathbf{b}-\mathbf{a})\)	✓ 1
A1: \(\frac{3}{4}\mathbf{a}+\frac{1}{4}\mathbf{b}\)	✓ 1
Part (c)
M1: \(\overrightarrow{QP}=\overrightarrow{OP}-\overrightarrow{OQ}=\frac{3}{4}\mathbf{a}+\frac{1}{4}\mathbf{b}-\frac{1}{4}\mathbf{b}\)	✓ 1
A1: \(\overrightarrow{QP}=\frac{3}{4}\mathbf{a}\), a multiple of \(\mathbf{a}\), so QP is parallel to OA	✓ 1

Common error: A ratio of 1 : 3 splits AB into 4 equal parts, so AP is \(\frac{1}{4}\) of AB, not \(\frac{1}{3}\).

+ Add to test✓ In your test19IGMC-047-01-08Vectors, magnitude and vector geometry Paper EMedium5 marks

Calculator

\(\mathbf{v}=\begin{pmatrix}5\\-3\end{pmatrix}\) and \(\mathbf{w}=2\mathbf{v}+\begin{pmatrix}-4\\9\end{pmatrix}\).

(a)

Show that \(\mathbf{w}=\begin{pmatrix}6\\3\end{pmatrix}\).

(1)

(b)

Find \(|\mathbf{w}|\), giving your answer correct to 3 significant figures.

(2)

(c)

Find the angle between \(\mathbf{w}\) and the positive \(x\)-direction, giving your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: \(\begin{pmatrix}10\\-6\end{pmatrix}+\begin{pmatrix}-4\\9\end{pmatrix}=\begin{pmatrix}6\\3\end{pmatrix}\) shown	✓ 1
Part (b)
M1: \(\sqrt{6^2+3^2}=\sqrt{45}\)	✓ 1
A1: 6.71	✓ 1
Part (c)
M1: \(\tan\theta=\frac{3}{6}\), or equivalent such as \(\cos\theta=\frac{6}{\sqrt{45}}\) or \(\sin\theta=\frac{3}{\sqrt{45}}\)	✓ 1
A1: 26.6°	✓ 1

Exam tip: Sketch a quick right-angled triangle from the components: horizontal 6, vertical 3. The angle with the \(x\)-direction then follows from \(\tan\theta=\frac{\text{vertical}}{\text{horizontal}}\).

+ Add to test✓ In your test20IGMC-047-01-09Vectors, magnitude and vector geometry Paper EMedium6 marks

Calculator

O is the origin. A is the point (1, 2) and B is the point (7, 10). P lies on AB such that AP : PB = 3 : 1.

(a)

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

(1)

(b)

Find \(|\overrightarrow{AB}|\).

(1)

(c)

Find the position vector of P.

(2)

(d)

Find \(|\overrightarrow{OP}|\), giving your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: \(\begin{pmatrix}6\\8\end{pmatrix}\)	✓ 1
Part (b)
B1: 10	✓ 1
Part (c)
M1: \(\begin{pmatrix}1\\2\end{pmatrix}+\frac{3}{4}\begin{pmatrix}6\\8\end{pmatrix}\)	✓ 1
A1: \(\begin{pmatrix}5.5\\8\end{pmatrix}\)	✓ 1
Part (d)
M1: \(\sqrt{5.5^2+8^2}=\sqrt{94.25}\)	✓ 1
A1: 9.71	✓ 1

Common error: \(\frac{3}{4}\overrightarrow{AB}\) is the vector from A to P, not the position of P. Add it to the position vector of A.

+ Add to test✓ In your test21IGMC-047-01-10Vectors, magnitude and vector geometry Paper EMedium5 marks

No calculator

\(\mathbf{a}=\begin{pmatrix}2\\1\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}-1\\3\end{pmatrix}\).

(a)

Find the values of \(m\) and \(n\) such that \(m\mathbf{a}+n\mathbf{b}=\begin{pmatrix}7\\7\end{pmatrix}\).

(4)

(b)

Find \(\left|\begin{pmatrix}7\\7\end{pmatrix}\right|\), giving your answer as a simplified surd.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: \(2m-n=7\) and \(m+3n=7\) formed	✓ 1
M1: correct method to eliminate one variable, e.g. \(n=2m-7\) substituted to give \(7m-21=7\)	✓ 1
A1: \(m=4\)	✓ 1
A1: \(n=1\)	✓ 1
Part (b)
B1: \(7\sqrt{2}\)	✓ 1

Exam tip: Each row of the column vectors gives one equation. Check your values by substituting back: \(4\mathbf{a}+\mathbf{b}=\begin{pmatrix}7\\7\end{pmatrix}\).

+ Add to test✓ In your test22IGMC-047-01-11Vectors, magnitude and vector geometry Paper EMedium6 marks

Calculator

ABCDEF is a regular hexagon with centre O. \(\overrightarrow{OA}=\mathbf{a}\) and \(\overrightarrow{OB}=\mathbf{b}\).

(a)

Find \(\overrightarrow{AB}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).

(1)

(b)

Explain why \(\overrightarrow{OC}=\mathbf{b}-\mathbf{a}\).

(2)

(c)

Find \(\overrightarrow{AD}\) in terms of \(\mathbf{a}\).

(1)

(d)

\(|\mathbf{a}|=6\) cm. Find \(|\overrightarrow{AC}|\), giving your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: \(\mathbf{b}-\mathbf{a}\)	✓ 1
Part (b)
M1: OABC is a rhombus (triangles OAB and OBC are equilateral), so OC is parallel and equal to AB	✓ 1
A1: hence \(\overrightarrow{OC}=\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\)	✓ 1
Part (c)
B1: \(-2\mathbf{a}\)	✓ 1
Part (d)
M1: \(AC^2=6^2+6^2-2\times6\times6\cos120°\) or \(2\times6\sin60°\)	✓ 1
A1: 10.4 cm	✓ 1

Exam tip: A regular hexagon splits into six equilateral triangles about its centre, so every side and every radius has the same length and interior angles are 120°.

+ Add to test✓ In your test23IGMC-047-00-11Reflections, rotations, translations and enlargements described in words Paper EHard5 marks

No calculator

Triangle A has vertices (0, 4), (2, 4) and (2, 6).
Triangle B has vertices (3, 1), (3, 3) and (5, 3).
(0, 4) maps to (3, 1), (2, 4) maps to (3, 3) and (2, 6) maps to (5, 3).

(a)

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

(2)

(b)

Find the image of the point (6, −2) under this transformation.

(2)

(c)

Write down the coordinates of the point on the x-axis that is mapped onto itself by this transformation.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for reflection	✓ 1
B1 for in the line y = x + 1	✓ 1
Part (b)
M1 for using the midpoint on y = x + 1 with a perpendicular joining line, or the rule (x, y) → (y − 1, x + 1)	✓ 1
A1 for (−3, 7)	✓ 1
Part (c)
B1 for (−1, 0)	✓ 1

Exam tip: The mirror line passes through the midpoint of each point and its image, and it is perpendicular to the line joining them. The midpoint of (0, 4) and (3, 1) is (1.5, 2.5), which lies on y = x + 1.

+ Add to test✓ In your test24IGMC-047-00-13Reflections, rotations, translations and enlargements described in words Paper EHard7 marks

No calculator

Rectangle R has vertices (−4, 2), (8, 2), (8, 8) and (−4, 8).
R is enlarged with scale factor \(\frac{1}{3}\) and centre (2, −1) to give rectangle S.

(a)

Find the coordinates of the vertices of S.

(3)

(b)

Describe fully the single transformation that maps S onto R.

(2)

(c)

Find the area of S.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(0, 0), (4, 0), (4, 2), (0, 2)	—
M1 for a correct method, e.g. (−4 − 2, 2 − (−1)) = (−6, 3) multiplied by \(\frac{1}{3}\) and added to (2, −1)	✓ 1
A1 for two vertices correct	✓ 1
A1 for all four vertices correct	✓ 1
Part (b)
B1 for enlargement with centre (2, −1)	✓ 1
B1 for scale factor 3	✓ 1
Part (c)
M1 for 12 × 6 × \(\left(\frac{1}{3}\right)^2\), or for 4 × 2 using their S	✓ 1
A1 for 8 square units	✓ 1

Exam tip: For a fractional scale factor, the image lies between the centre and the original shape. Multiply each displacement from the centre by the scale factor. Do not multiply the coordinates themselves unless the centre is the origin.

+ Add to test✓ In your test25IGMC-047-00-14Reflections, rotations, translations and enlargements described in words Paper EHard7 marks

Calculator

A rotation of 180° about the point C maps the point P(3, −1) onto the point P′(−5, 7).

(a)

Find the coordinates of C.

(2)

(b)

Find the image of the point Q(0, 4) under the same rotation.

(2)

(c)

A translation also maps P onto P′.
Write down the vector of this translation and find its magnitude. Give the magnitude to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(\left(\frac{3 + (-5)}{2}, \frac{-1 + 7}{2}\right)\)	✓ 1
A1 for (−1, 3)	✓ 1
Part (b)
M1 for using C as the midpoint of Q and its image, e.g. (2 × (−1) − 0, 2 × 3 − 4)	✓ 1
A1 for (−2, 2)	✓ 1
Part (c)
B1 for \(\begin{pmatrix}-8\\8\end{pmatrix}\)	✓ 1
M1 for \(\sqrt{(-8)^2 + 8^2}\) or \(\sqrt{128}\)	✓ 1
A1 for 11.3 units. An answer left only as \(8\sqrt{2}\) does not earn this mark, because it is not given to 3 s.f.	✓ 1

Common error: Squaring a negative component on a calculator without brackets, as in −8², gives −64 instead of 64. Always type (−8)² when finding a magnitude.

+ Add to test✓ In your test26IGMC-047-01-12Vectors, magnitude and vector geometry Paper EHard8 marks

No calculator

In triangle OAB, \(\overrightarrow{OA}=\mathbf{a}\) and \(\overrightarrow{OB}=\mathbf{b}\). M is the midpoint of OA. N lies on AB such that AN : NB = 2 : 1. The line MN is extended to meet the line OB extended at X, where \(\overrightarrow{OX}=k\mathbf{b}\).

(a)

Find \(\overrightarrow{AB}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).

(1)

(b)

Find \(\overrightarrow{ON}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\), in its simplest form.

(2)

(c)

Find \(\overrightarrow{MN}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\), in its simplest form.

(2)

(d)

Find the value of \(k\).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: \(\mathbf{b}-\mathbf{a}\)	✓ 1
Part (b)
M1: \(\mathbf{a}+\frac{2}{3}(\mathbf{b}-\mathbf{a})\)	✓ 1
A1: \(\frac{1}{3}\mathbf{a}+\frac{2}{3}\mathbf{b}\)	✓ 1
Part (c)
M1: \(\overrightarrow{ON}-\frac{1}{2}\mathbf{a}\)	✓ 1
A1: \(-\frac{1}{6}\mathbf{a}+\frac{2}{3}\mathbf{b}\)	✓ 1
Part (d)
M1: \(\overrightarrow{MX}=k\mathbf{b}-\frac{1}{2}\mathbf{a}\)	✓ 1
M1: \(\overrightarrow{MX}=\lambda\overrightarrow{MN}\) and comparing coefficients of \(\mathbf{a}\): \(-\frac{1}{2}=-\frac{\lambda}{6}\), so \(\lambda=3\)	✓ 1
A1: \(k=3\times\frac{2}{3}=2\)	✓ 1

Exam tip: For points on a straight line, write one vector as a scalar multiple of another, then match the coefficients of \(\mathbf{a}\) and of \(\mathbf{b}\) separately.

+ Add to test✓ In your test27IGMC-047-01-13Vectors, magnitude and vector geometry Paper EHard7 marks

Calculator

Distances are in kilometres. At 12:00 a boat is at the point with position vector \(\begin{pmatrix}2\\3\end{pmatrix}\). It moves with constant velocity \(\begin{pmatrix}4\\-1\end{pmatrix}\) km/h. A lighthouse L is at the point with position vector \(\begin{pmatrix}10\\5\end{pmatrix}\). The vector \(\begin{pmatrix}0\\1\end{pmatrix}\) points due north.

(a)

Write down the position vector of the boat \(t\) hours after 12:00.

(2)

(b)

Find the time when the boat is due south of L, and its distance from L at that time.

(3)

(c)

Find the distance of the boat from L at 15:00, giving your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: \(x\)-component \(2+4t\)	✓ 1
B1: \(y\)-component \(3-t\)	✓ 1
Part (b)
M1: \(2+4t=10\)	✓ 1
A1: \(t=2\), so 14:00	✓ 1
A1: boat at (10, 1), distance 4 km	✓ 1
Part (c)
M1: boat at \(\begin{pmatrix}14\\0\end{pmatrix}\), vector from L \(\begin{pmatrix}4\\-5\end{pmatrix}\) and \(\sqrt{4^2+5^2}\)	✓ 1
A1: 6.40 km	✓ 1

Common error: Position after \(t\) hours is the start position plus \(t\) times the velocity. Adding the velocity only once gives the position after 1 hour, whatever the value of \(t\).

+ Add to test✓ In your test28IGMC-047-01-14Vectors, magnitude and vector geometry Paper EHard7 marks

No calculator

OABC is a trapezium. \(\overrightarrow{OA}=\mathbf{a}\), \(\overrightarrow{OC}=\mathbf{c}\) and \(\overrightarrow{CB}=2\mathbf{a}\). The diagonals OB and AC meet at X.

(a)

Find, in terms of \(\mathbf{a}\) and \(\mathbf{c}\), (i) \(\overrightarrow{OB}\) (ii) \(\overrightarrow{AC}\).

(2)

(b)

\(\overrightarrow{OX}=\mu\overrightarrow{OB}\) and \(\overrightarrow{AX}=\lambda\overrightarrow{AC}\). Find \(\overrightarrow{OX}\) in terms of \(\mathbf{a}\) and \(\mathbf{c}\).

(4)

(c)

Write down the ratio OX : XB.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: (i) \(2\mathbf{a}+\mathbf{c}\)	✓ 1
B1: (ii) \(\mathbf{c}-\mathbf{a}\)	✓ 1
Part (b)
M1: \(\overrightarrow{OX}=\mathbf{a}+\lambda(\mathbf{c}-\mathbf{a})=(1-\lambda)\mathbf{a}+\lambda\mathbf{c}\)	✓ 1
M1: equating with \(2\mu\mathbf{a}+\mu\mathbf{c}\): \(2\mu=1-\lambda\) and \(\mu=\lambda\)	✓ 1
A1: \(\mu=\lambda=\frac{1}{3}\)	✓ 1
A1: \(\overrightarrow{OX}=\frac{2}{3}\mathbf{a}+\frac{1}{3}\mathbf{c}\)	✓ 1
Part (c)
B1: 1 : 2	✓ 1

Exam tip: When two lines cross, write the crossing point in two different ways using two unknown scalars, then compare coefficients because \(\mathbf{a}\) and \(\mathbf{c}\) are not parallel.