International GCSE Maths · Pearson Edexcel 4MA1 · Unit 5
Vectors and transformation geometry
54 free, high-quality past-paper-style questions on Vectors and transformation geometry for Edexcel IGCSE Maths, each with a full mark scheme. Covers: Transformations described in words; Vectors and vector geometry. For when you have finished the past papers, or want to practise one topic at a time.Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.283 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IGME-062-01-02Transformations described in wordsPaper FEasy4 marks
The point P has coordinates (5, 3).
Diagram on a coordinate grid
(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 point
Mark
Notes
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 wordsPaper 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 point
Mark
Notes
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 wordsPaper 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.
Diagram on a coordinate grid
(a)
Write down the coordinates of the vertices of triangle B.
(2)
(b)
Find the area of triangle B.
(2)
Show mark scheme
Marking point
Mark
Notes
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 wordsPaper 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).
Diagram on a coordinate grid
(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 point
Mark
Notes
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 geometryPaper 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 point
Mark
Notes
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 geometryPaper FEasy3 marks
The point A has coordinates (2, 7) and the point B has coordinates (6, −1).
Diagram on a coordinate grid
(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 point
Mark
Notes
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 geometryPaper FEasy3 marks
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-062-00-05Transformations described in wordsPaper 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.
Diagram on a coordinate grid
(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 point
Mark
Notes
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.
9IGME3-5-12r39-01Transformations described in wordsPaper FEasy5 marks
A map of a wind farm uses a coordinate grid, with 1 unit representing 100 m. A turbine stands at A(2, 3). It is replaced by a turbine at A′, found by translating A by the vector \(\begin{pmatrix}5\\-2\end{pmatrix}\).
Diagram on a coordinate grid
(a)
Write down the coordinates of A′.
(2)
(b)
Another turbine at B(−1, 4) is moved to B′(3, 1). Write down the column vector of this translation.
(2)
(c)
Write down the column vector of the translation that moves A′ back to A.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (2 + 5, 3 − 2)
✓ 1
A1 (7, 1)
✓ 1
Part (b)
M1 3 − (−1) and 1 − 4
✓ 1
A1 \(\begin{pmatrix}4\\-3\end{pmatrix}\)
✓ 1
Part (c)
B1 \(\begin{pmatrix}-5\\2\end{pmatrix}\)
✓ 1
Common error: In a column vector the top number is the horizontal movement and the bottom number is the vertical movement; subtract start from end, not the other way round.
10IGME3-5-12r39-02Transformations described in wordsPaper FEasy6 marks
A shop plan is drawn on a coordinate grid. A display stand is at P(3, 2).
Diagram on a coordinate grid
(a)
P is reflected in the line x = 5. Write down the coordinates of the image of P.
(2)
(b)
A stand at Q(6, −1) is reflected in the line y = 1. Write down the coordinates of the image of Q.
(2)
(c)
A triangle with vertices (1, 1), (3, 1) and (1, 4) is mapped to a triangle with vertices (1, −1), (3, −1) and (1, −4). Describe fully the single transformation.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P is 2 units to the left of the line so the image is 2 units to the right
✓ 1
A1 (7, 2)
✓ 1
Part (b)
M1 Q is 2 units below the line so the image is 2 units above
✓ 1
A1 (6, 3)
✓ 1
Part (c)
B1 reflection
✓ 1
B1 in the x-axis (the line y = 0)
✓ 1
Exam tip: A point and its mirror image are the same distance from the mirror line, measured at right angles to it.
11IGME3-5-12r39-03Transformations described in wordsPaper FEasy5 marks
A health agency plots testing sites on a coordinate grid. The sites are rotated on a new planning map. All rotations in this question are about the point stated.
Diagram on a coordinate grid
(a)
Site A(2, 1) is rotated through 90° clockwise about the origin. Write down the coordinates of its image.
(2)
(b)
Site B(−3, 2) is rotated through 180° about the origin. Write down the coordinates of its image.
(1)
(c)
Site C(4, 1) is rotated through 90° anticlockwise about the point (1, 1). Write down the coordinates of its image.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 a rotation by 90° of the point, for example (−1, 2) or (1, −2) seen
✓ 1
A1 (1, −2)
✓ 1
Part (b)
B1 (3, −2)
✓ 1
Part (c)
M1 C is 3 units to the right of the centre so image is 3 units above the centre
✓ 1
A1 (1, 4)
✓ 1
Common error: Check the direction of the turn and the centre of rotation; a quarter turn about a point other than the origin should be done relative to that point.
12IGME3-5-12r39-04Transformations described in wordsPaper FEasy4 marks
On a coordinate grid, a shop designer enlarges displays. All the enlargements below have a positive scale factor.
(a)
The point (2, 1) is enlarged by scale factor 3 with centre the origin. Write down the coordinates of the image.
(1)
(b)
The point (3, 2) is enlarged by scale factor 2 with centre (1, 1). Write down the coordinates of the image.
(2)
(c)
A poster has a side of length 6 cm. In an enlargement the image of this side has length 15 cm. Work out the scale factor.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 (6, 3)
✓ 1
Part (b)
M1 vector from the centre to the point is \(\begin{pmatrix}2\\1\end{pmatrix}\) which is doubled to \(\begin{pmatrix}4\\2\end{pmatrix}\)
✓ 1
A1 (5, 3)
✓ 1
Part (c)
B1 2.5
✓ 1
Exam tip: For an enlargement from a centre that is not the origin, measure from the centre to the point, multiply that vector by the scale factor, then add it back to the centre.
13IGME3-5-13r45-01Transformations described in wordsPaper FEasy4 marks
A marine biologist records a seagrass survey plot on a grid where 1 unit represents 1 m. The plot is triangle ABC with A(1, 2), B(4, 2) and C(4, 6). The plot is translated by the vector \(\begin{pmatrix}-5\\3\end{pmatrix}\) to give triangle A'B'C'.
Diagram on a coordinate grid
(a)
Write down the coordinates of A', B' and C'.
(2)
(b)
Work out the perimeter of triangle A'B'C'.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 two image points correct
✓ 1
B1 A'(−4, 5), B'(−1, 5), C'(−1, 9)
✓ 1
Part (b)
M1 recognises the translation preserves lengths and finds AC = \(\sqrt{3^2+4^2}\) = 5 (or finds a length of the image directly)
✓ 1
A1 3 + 4 + 5 = 12 m
✓ 1
Exam tip: A translation changes position only, so lengths and angles in the image match the original. Use that to save work.
14IGME3-5-13r45-02Transformations described in wordsPaper FEasy4 marks
On an architect's floor plan drawn on a coordinate grid, triangle ABC has A(2, 1), B(2, 4) and C(4, 1). It is mapped onto triangle A'B'C' where A' is (−2, 1), B' is (−2, 4) and C' is (−4, 1).
Diagram on a coordinate grid
(a)
Describe fully the single transformation that maps triangle ABC onto triangle A'B'C'.
(2)
(b)
The point P(3, −2) is reflected in the line y = 1. Write down the coordinates of the image of P.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 reflection
✓ 1
B1 in the y-axis (the line x = 0)
✓ 1
Part (b)
M1 P is 3 units below the line y = 1, so the image is 3 units above it (or 1 + 3)
✓ 1
A1 (3, 4)
✓ 1
Common error: When reflecting in a horizontal or vertical line other than an axis, measure the distance from the point to the mirror line, not to the axis.
15IGME3-5-13r45-03Transformations described in wordsPaper FEasy5 marks
A warehouse floor plan is drawn on a grid. A pallet marker is the triangle PQR with P(1, 1), Q(3, 1) and R(3, 2). Triangle PQR is rotated 90° clockwise about the origin to give triangle P'Q'R'.
Diagram on a coordinate grid
(a)
Write down the coordinates of P', Q' and R'.
(2)
(b)
Describe fully the single transformation that maps triangle P'Q'R' onto triangle PQR.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 two image points correct
✓ 1
B1 P'(1, −1), Q'(1, −3), R'(2, −3)
✓ 1
Part (b)
B1 rotation
✓ 1
B1 90° anticlockwise
✓ 1
B1 about (0, 0)
✓ 1
Exam tip: A rotation needs three details: the angle, the direction and the centre. Trace the point with a sketch if unsure.
16IGME3-5-13r45-04Transformations described in wordsPaper FEasy4 marks
A coral monitoring plot is drawn on a grid where 1 unit represents 1 m. The plot is the rectangle with vertices (1, 1), (4, 1), (4, 3) and (1, 3). It is enlarged by scale factor 3 with centre (0, 0).
Diagram on a coordinate grid
(a)
Write down the coordinates of the image of the vertex (4, 3).
(1)
(b)
The area of the original plot is 1 part and the area of the image is n parts. Write down the value of n.
(1)
(c)
Work out the area of the image of the plot.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 (12, 9)
✓ 1
Part (b)
B1 9
✓ 1
Part (c)
M1 lengths of the image are 9 m and 6 m (or 6 × their ratio from part (b))
✓ 1
A1 54 m2
✓ 1
Common error: Lengths are multiplied by the scale factor, but areas are multiplied by the scale factor squared.
17IGME-062-01-06Transformations described in wordsPaper 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.
Diagram on a coordinate grid
(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 point
Mark
Notes
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.
18IGME-062-01-07Transformations described in wordsPaper 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.
Diagram on a coordinate grid
(a)
Find the coordinates of the vertices of triangle S.
(3)
(b)
Find the area of triangle S.
(2)
Show mark scheme
Marking point
Mark
Notes
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).
19IGME-062-01-08Transformations described in wordsPaper 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.
Diagram on a coordinate grid
(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 point
Mark
Notes
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).
20IGME-062-01-10Transformations described in wordsPaper 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.
Diagram on a coordinate grid
(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 point
Mark
Notes
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.
21IGME-062-01-11Transformations described in wordsPaper FMedium4 marks
A translation maps the point A(−3, 5) onto the point A′(4, −1).
Diagram on a coordinate grid
(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 point
Mark
Notes
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.
22IGME-062-00-08Vectors and vector geometryPaper 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 point
Mark
Notes
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.
23IGME3-5-12r39-05Transformations described in wordsPaper FMedium5 marks
Three triangles are drawn on a coordinate grid to plan solar panel layouts. Triangle T has vertices (1, 1), (2, 1) and (1, 3). Triangle U has vertices (2, 2), (4, 2) and (2, 6). Triangle V has vertices (−1, 1), (−2, 1) and (−1, 3).
Diagram on a coordinate grid
(a)
Describe fully the single transformation that maps T onto U.
(3)
(b)
Describe fully the single transformation that maps T onto V.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 enlargement
✓ 1
B1 scale factor 2
✓ 1
B1 centre (0, 0)
✓ 1
Part (b)
B1 reflection
✓ 1
B1 in the y-axis (the line x = 0)
✓ 1
Common error: An enlargement needs both a scale factor and a centre; if either is missing the description is incomplete.
24IGME3-5-12r39-06Vectors and vector geometryPaper FMedium5 marks
A drone inspects a solar farm. The grid uses 1 unit for 1 km. The drone leaves the base O, flies a first leg with displacement \(\begin{pmatrix}5\\2\end{pmatrix}\) and then a second leg with displacement \(\begin{pmatrix}-3\\4\end{pmatrix}\).
Sketch of the route, not to scale
(a)
Write down the total displacement of the drone from O after the two legs.
(2)
(b)
Calculate the straight-line distance of the drone from O after the two legs. Give your answer correct to 3 significant figures.
(2)
(c)
The drone now flies directly back to O. Write down the displacement for this journey.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (5 + (−3), 2 + 4)
✓ 1
A1 \(\begin{pmatrix}2\\6\end{pmatrix}\)
✓ 1
Part (b)
M1 √(22 + 62) (ECF from (a))
✓ 1
A1 6.32 km
✓ 1
Part (c)
B1 \(\begin{pmatrix}-2\\-6\end{pmatrix}\) (ECF from (a))
✓ 1
Exam tip: The magnitude of a vector is found with Pythagoras' theorem on its two components; the direction does not matter.
25IGME3-5-12r39-07Transformations described in wordsPaper FMedium6 marks
A point P(2, 3) marks the corner of a solar panel on a coordinate grid. The corner is moved by two transformations, one after the other.
(a)
P is translated by the vector \(\begin{pmatrix}-4\\1\end{pmatrix}\). Write down the coordinates of the image, P′.
(2)
(b)
P′ is then reflected in the line y = x. Write down the coordinates of the image, P″.
(2)
(c)
A panel is moved by the same two transformations. Is the final panel congruent to the original? Give a reason for your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (2 − 4, 3 + 1)
✓ 1
A1 (−2, 4)
✓ 1
Part (b)
M1 the x and y coordinates are swapped (ECF from (a))
✓ 1
A1 (4, −2)
✓ 1
Part (c)
B1 yes
✓ 1
B1 translations and reflections do not change lengths or angles
✓ 1
Common error: Reflection in y = x swaps the coordinates; reflection in y = −x swaps them and changes both signs.
26IGME3-5-12r39-08Transformations described in wordsPaper FMedium5 marks
A retailer places floor stickers in the shape of triangle PQR on a coordinate grid, with P(2, 1), Q(4, 1) and R(2, 3). The triangle is enlarged with scale factor 2 and centre (1, 0) to give triangle P′Q′R′.
Diagram on a coordinate grid
(a)
Write down the coordinates of P′.
(2)
(b)
Write down the coordinates of R′.
(2)
(c)
The length of PQ is 2 units. Write down the length of P′Q′.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 vector from the centre to P is \(\begin{pmatrix}1\\1\end{pmatrix}\), doubled to \(\begin{pmatrix}2\\2\end{pmatrix}\)
✓ 1
A1 (3, 2)
✓ 1
Part (b)
M1 vector from the centre to R is \(\begin{pmatrix}1\\3\end{pmatrix}\), doubled to \(\begin{pmatrix}2\\6\end{pmatrix}\)
✓ 1
A1 (3, 6)
✓ 1
Part (c)
B1 4
✓ 1
Exam tip: Every length in an enlargement is multiplied by the scale factor, and so is every distance measured from the centre of enlargement.
27IGME3-5-13r45-08Transformations described in wordsPaper FMedium7 marks
An architect makes a scale model. On a coordinate grid (1 unit = 1 m) the triangular base of the real building is triangle PQR with P(2, 2), Q(8, 2) and R(8, 6). The base is enlarged by scale factor \(\frac{1}{2}\) with centre (0, 0) to give the model base P'Q'R'.
Diagram on a coordinate grid
(a)
Write down the coordinates of P', Q' and R'.
(2)
(b)
Describe fully the single transformation that maps triangle P'Q'R' onto triangle PQR.
(3)
(c)
Work out the area of triangle P'Q'R'.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 two image points correct
✓ 1
B1 P'(1, 1), Q'(4, 1), R'(4, 3)
✓ 1
Part (b)
B1 enlargement
✓ 1
B1 scale factor 2
✓ 1
B1 centre (0, 0)
✓ 1
Part (c)
M1 \(\frac{1}{2}\times 3\times 2\)
✓ 1
A1 3 m2
✓ 1
Common error: A scale factor between 0 and 1 makes the image smaller. To reverse an enlargement, use the reciprocal of the scale factor with the same centre.
28IGME3-5-13r45-09Transformations described in wordsPaper FMedium5 marks
A logistics planner moves a crate on a grid where 1 unit represents 1 m. Triangle T is the crate's footprint. It is moved to triangle U by the translation \(\begin{pmatrix}6\\-4\end{pmatrix}\) and then to triangle V by the translation \(\begin{pmatrix}-10\\1\end{pmatrix}\).
(a)
Write down the single translation, as a column vector, that maps T directly onto V.
(2)
(b)
The point (5, 5) is a corner of T. Write down the coordinates of the corresponding corner of V.
(1)
(c)
Work out the straight-line distance the crate has moved from T to V.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 adding the two vectors (at least one component correct)
✓ 1
A1 \(\begin{pmatrix}-4\\-3\end{pmatrix}\)
✓ 1
Part (b)
B1 (1, 2) (ECF from (a))
✓ 1
Part (c)
M1 \(\sqrt{4^2+3^2}\)
✓ 1
A1 5 m
✓ 1
Exam tip: Two translations in a row combine by adding their column vectors. The distance moved is the length of the combined vector.
29IGME3-5-13r45-12Transformations described in wordsPaper FMedium5 marks
A warehouse robot travels between two racks, A(1, 2) and B(7, 10), on a coordinate grid where 1 unit represents 1 m. A charging point M is placed at the midpoint of AB.
Diagram on a coordinate grid
(a)
Find the coordinates of M.
(2)
(b)
Write down the translation, as a column vector, that maps A onto B.
Exam tip: The midpoint is the mean of the x-coordinates and the mean of the y-coordinates. The length comes from Pythagoras on the column vector.
30IGME-062-01-12Transformations described in wordsPaper 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.
Diagram on a coordinate grid
(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 point
Mark
Notes
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.
31IGME3-5-13r45-07Transformations described in wordsPaper FHard7 marks
On an architect's plan grid, a triangular roof panel has vertices A(1, 2), B(3, 2) and C(3, 5). Triangle ABC is rotated 180° about the origin to give triangle A1B1C1. Triangle A1B1C1 is then translated by \(\begin{pmatrix}4\\7\end{pmatrix}\) to give triangle A2B2C2.
Diagram on a coordinate grid
(a)
Write down the coordinates of A1, B1 and C1.
(2)
(b)
Write down the coordinates of A2, B2 and C2.
(2)
(c)
Describe fully the single transformation that maps triangle ABC directly onto triangle A2B2C2.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 two image points correct
✓ 1
B1 A1(−1, −2), B1(−3, −2), C1(−3, −5)
✓ 1
Part (b)
B1 two image points correct (ECF from (a))
✓ 1
B1 A2(3, 5), B2(1, 5), C2(1, 2)
✓ 1
Part (c)
B1 rotation (half turn)
✓ 1
B1 180°
✓ 1
B1 about the point (2, 3.5)
✓ 1
Exam tip: The centre of a half turn is the midpoint between any point and its image. Check with a second point.
32IGME-062-00-03Vectors and vector geometryPaper 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 point
Mark
Notes
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.
33IGME-062-01-09Transformations described in wordsPaper 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.
Diagram on a coordinate grid
(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 point
Mark
Notes
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.
34IGME-062-00-06Vectors and vector geometryPaper 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.
Diagram not to scale
(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 point
Mark
Notes
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.
35IGME-062-00-09Vectors and vector geometryPaper HMedium4 marks
OABC is a parallelogram with vertices in that order. \(\overrightarrow{OA}\) = a and \(\overrightarrow{OC}\) = c.
Diagram not to scale
(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 point
Mark
Notes
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.
36IGME-062-00-10Vectors and vector geometryPaper 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 \(\begin{pmatrix}-4\\10\end{pmatrix}\).
Sketch of the route, not to scale
(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 point
Mark
Notes
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.
37IGME-062-00-11Vectors and vector geometryPaper 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.
Diagram on a coordinate grid
(a)
Use vectors to find the coordinates of M.
(2)
(b)
Find the length of AM.
(2)
Show mark scheme
Marking point
Mark
Notes
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.
38IGME-062-00-07Vectors and vector geometryPaper 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 point
Mark
Notes
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°.
39IGME3-5-12r39-09Vectors and vector geometryPaper HMedium5 marks
Cables in a wind farm run from a substation O to two turbines A and B, where OA = a and OB = b. The point M is the midpoint of AB. The point N lies on OB such that ON = \(\frac{1}{3}\)OB.
Diagram not to scale
(a)
Write down AB in terms of a and b.
(1)
(b)
Find OM in terms of a and b. Give your answer in its simplest form.
(2)
(c)
Find NM in terms of a and b. Give your answer in its simplest form.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 b − a
✓ 1
Part (b)
M1 OA + \(\frac{1}{2}\)AB or equivalent
✓ 1
A1 \(\frac{1}{2}\)(a + b)
✓ 1
Part (c)
M1 OM − ON with ON = \(\frac{1}{3}\)b (ECF from (b))
✓ 1
A1 \(\frac{1}{2}\)a + \(\frac{1}{6}\)b
✓ 1
Exam tip: To reach any point, choose a route made only of vectors you already know, and simplify by collecting the a terms and the b terms separately.
40IGME3-5-12r39-11Transformations described in wordsPaper HMedium6 marks
On a coordinate grid with 1 unit = 1 cm, a retailer designs a logo. Triangle T has vertices (2, 2), (6, 2) and (2, 4). Triangle U has vertices (−1, −1), (−3, −1) and (−1, −2).
Diagram on a coordinate grid
(a)
Describe fully the single transformation that maps T onto U.
(3)
(b)
Work out the area of triangle U by using the area of triangle T and the scale factor of the enlargement.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 enlargement
✓ 1
B1 scale factor −\(\frac{1}{2}\)
✓ 1
B1 centre (0, 0)
✓ 1
Part (b)
M1 area of T = \(\frac{1}{2}\) × 4 × 2 = 4 cm2
✓ 1
M1 4 × (−\(\frac{1}{2}\))2 or 4 × \(\frac{1}{4}\)
✓ 1
A1 1 cm2
✓ 1
Common error: Areas change by the square of the scale factor, so a negative scale factor still gives a positive area factor.
41IGME3-5-12r39-14Vectors and vector geometryPaper HMedium7 marks
Two vaccination sites are at A(2, 3) and B(8, −3) on a coordinate grid, where 1 unit represents 1 km. A mobile clinic stands at P on the line AB such that AP : PB = 1 : 2.
Diagram on a coordinate grid
(a)
Write down the vector AB as a column vector.
(1)
(b)
Find the coordinates of P.
(2)
(c)
A second clinic is at Q, on the line AB extended beyond B, with BQ = \(\frac{1}{2}\)AB. Find the coordinates of Q.
(2)
(d)
Calculate the distance AB. Give your answer correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 \(\begin{pmatrix}6\\-6\end{pmatrix}\)
✓ 1
Part (b)
M1 AP = \(\frac{1}{3}\)AB = \(\begin{pmatrix}2\\-2\end{pmatrix}\)
✓ 1
A1 (4, 1)
✓ 1
Part (c)
M1 OB + \(\frac{1}{2}\)AB or (8 + 3, −3 − 3)
✓ 1
A1 (11, −6)
✓ 1
Part (d)
M1 √(62 + 62)
✓ 1
A1 8.49 km
✓ 1
Common error: A ratio of 1 : 2 divides the line into three equal parts, so the first part is one third of the whole line, not one half.
42IGME3-5-13r45-05Vectors and vector geometryPaper HMedium6 marks
In a logistics model, the displacements (in metres) of a crane trolley on two moves are given by the vectors \(\mathbf{a}=\begin{pmatrix}3\\-2\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}-1\\4\end{pmatrix}\).
(a)
Work out \(2\mathbf{a}+3\mathbf{b}\) as a column vector.
(2)
(b)
The vector \(\mathbf{a}+k\mathbf{b}\) is horizontal. Find the value of k.
(2)
(c)
Work out the magnitude of \(2\mathbf{a}+3\mathbf{b}\), giving your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\begin{pmatrix}6\\-4\end{pmatrix}+\begin{pmatrix}-3\\12\end{pmatrix}\) (at least one component correct)
✓ 1
A1 \(\begin{pmatrix}3\\8\end{pmatrix}\)
✓ 1
Part (b)
M1 vertical component set to zero: −2 + 4k = 0
✓ 1
A1 k = 0.5
✓ 1
Part (c)
M1 \(\sqrt{3^2+8^2}\) (ECF from (a))
✓ 1
A1 8.54 m
✓ 1
Exam tip: A horizontal vector has zero vertical component. Setting that component equal to zero gives an equation to solve.
43IGME3-5-13r45-06Vectors and vector geometryPaper HMedium5 marks
A research vessel leaves harbour H and sails to point A, a displacement of \(\begin{pmatrix}12\\5\end{pmatrix}\) km. It then sails to point B, a further displacement of \(\begin{pmatrix}-7\\9\end{pmatrix}\) km. (The top number is km east and the bottom number is km north.)
Sketch of the route, not to scale
(a)
Write down the vector \(\overrightarrow{HB}\).
(1)
(b)
Work out the distance HB, giving your answer to 3 significant figures.
(2)
(c)
The vessel returns directly from B to H at a constant speed of 20 km/h. Work out the time taken in minutes, giving your answer to 3 significant figures. Use the unrounded value of HB.
Exam tip: Add column vectors component by component to find a resultant displacement, then use Pythagoras for its length.
44IGME3-5-13r45-10Vectors and vector geometryPaper HMedium5 marks
In a model of a roof frame, O is the apex, \(\overrightarrow{OA}=\mathbf{a}\) and \(\overrightarrow{OB}=\mathbf{b}\). The point M is the midpoint of AB. The point N lies on OB such that \(ON=\frac{1}{3}OB\).
Diagram not to scale
(a)
Write down \(\overrightarrow{AB}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
(1)
(b)
Find \(\overrightarrow{OM}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\). Give your answer in its simplest form.
(2)
(c)
Find \(\overrightarrow{MN}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
Common error: To go along a route, add the vectors in order and reverse the sign when you travel against an arrow, for example \(\overrightarrow{MO}=-\overrightarrow{OM}\).
45IGME3-5-13r45-13Vectors and vector geometryPaper HMedium5 marks
Two ocean currents have velocities (in km/h) given by the vectors \(\mathbf{p}=\begin{pmatrix}4\\1\end{pmatrix}\) and \(\mathbf{q}=\begin{pmatrix}-1\\5\end{pmatrix}\).
(a)
Work out \(3\mathbf{p}-2\mathbf{q}\) as a column vector.
(2)
(b)
The vector \(\mathbf{p}+k\mathbf{q}\) is parallel to \(\begin{pmatrix}1\\-2\end{pmatrix}\). Find the value of k.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\begin{pmatrix}12\\3\end{pmatrix}-\begin{pmatrix}-2\\10\end{pmatrix}\) (at least one component correct)
Common error: Parallel vectors are scalar multiples of each other, so compare components using a common multiplier. Do not simply equate the two components.
46IGME-062-00-12Vectors and vector geometryPaper 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.
Diagram not to scale
(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 point
Mark
Notes
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.
47IGME-062-00-13Vectors and vector geometryPaper 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.
Diagram not to scale
(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 point
Mark
Notes
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.
48IGME-062-00-14Vectors and vector geometryPaper HHard6 marks
ABCDEF is a regular hexagon with vertices in that order. \(\overrightarrow{AB}\) = p and \(\overrightarrow{BC}\) = q.
Diagram not to scale
(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 point
Mark
Notes
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.
49IGME-062-01-14Transformations described in wordsPaper 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 point
Mark
Notes
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.
50IGME3-5-12r39-10Vectors and vector geometryPaper HHard9 marks
Health workers record the displacements, in kilometres, of two outreach vehicles. Vehicle P has displacement p = \(\begin{pmatrix}6\\-8\end{pmatrix}\) and vehicle Q has displacement q = \(\begin{pmatrix}-3\\4\end{pmatrix}\).
(a)
Show that p is parallel to q.
(2)
(b)
Work out the magnitude of p.
(2)
(c)
The vector r = \(\begin{pmatrix}9\\y\end{pmatrix}\) is parallel to q. Find the value of y.
(2)
(d)
A third vehicle has displacement p + kq, where k > 0. The magnitude of this displacement is 15 km. Find the value of k.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 p = −2q
✓ 1
B1 one vector is a scalar multiple of the other, so they are parallel
✓ 1
Part (b)
M1 √(62 + (−8)2)
✓ 1
A1 10
✓ 1
Part (c)
M1 9 = −3 × (−3), so scalar multiple is −3
✓ 1
A1 y = −12
✓ 1
Part (d)
M1 p + kq = (k − 2)q or components (6 − 3k, 4k − 8)
Exam tip: Parallel vectors are scalar multiples of each other; spotting a common factor can turn a long calculation into a one-line equation.
51IGME3-5-12r39-12Vectors and vector geometryPaper HHard8 marks
A solar cable run is modelled with OA = a and OB = b. The point C lies on AB such that AC : CB = 1 : 3. The point D is such that OD = 3a + b.
Diagram not to scale
(a)
Find AC in terms of a and b.
(2)
(b)
Show that OC = \(\frac{3}{4}\)a + \(\frac{1}{4}\)b.
(2)
(c)
Show that O, C and D lie on a straight line.
(2)
(d)
Find the ratio OC : CD.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 AB = b − a
✓ 1
A1 \(\frac{1}{4}\)(b − a)
✓ 1
Part (b)
M1 OC = OA + AC
✓ 1
A1 a + \(\frac{1}{4}\)b − \(\frac{1}{4}\)a = \(\frac{3}{4}\)a + \(\frac{1}{4}\)b
✓ 1
Part (c)
M1 OD = 4OC
✓ 1
R1 OD is a multiple of OC and both start at O, so the points are collinear
✓ 1
Part (d)
M1 CD = OD − OC = 3OC
✓ 1
A1 1 : 3
✓ 1
Exam tip: To prove three points are collinear, show that two vectors between them are multiples of each other and that they share a common point.
52IGME3-5-12r39-13Transformations described in wordsPaper HHard8 marks
A map of a district uses a coordinate grid. Triangle T has vertices (2, 1), (5, 1) and (2, 3). T is reflected in the line x = 3 to give triangle U. U is then reflected in the line x = 6 to give triangle V.
Diagram on a coordinate grid
(a)
Write down the coordinates of the vertices of U.
(2)
(b)
Write down the coordinates of the vertices of V.
(2)
(c)
Describe fully the single transformation that maps T onto V.
(2)
(d)
Instead, T is reflected in the line x = 3 and then in the line x = 9. Write down the column vector of the single transformation that maps T onto the final image.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 at least one vertex reflected correctly
✓ 1
A1 (4, 1), (1, 1), (4, 3)
✓ 1
Part (b)
M1 at least one vertex of U reflected in x = 6 (ECF from (a))
✓ 1
A1 (8, 1), (11, 1), (8, 3)
✓ 1
Part (c)
B1 translation
✓ 1
B1 by the vector \(\begin{pmatrix}6\\0\end{pmatrix}\)
✓ 1
Part (d)
M1 2 × (9 − 3)
✓ 1
A1 \(\begin{pmatrix}12\\0\end{pmatrix}\)
✓ 1
Exam tip: Two reflections in parallel lines give a translation of twice the distance between the lines, at right angles to them.
53IGME3-5-13r45-11Vectors and vector geometryPaper HHard6 marks
Two buoys are positioned relative to a fixed point O in a marine reserve. \(\overrightarrow{OA}=\mathbf{a}\) and \(\overrightarrow{OB}=\mathbf{b}\). The point X lies on AB such that AX : XB = 2 : 3. A third buoy Y is positioned so that \(\overrightarrow{OY}=6\mathbf{a}+4\mathbf{b}\).
Diagram not to scale
(a)
Show that \(\overrightarrow{OX}=\frac{3}{5}\mathbf{a}+\frac{2}{5}\mathbf{b}\).
A1 \(\overrightarrow{OX}=\mathbf{a}+\frac{2}{5}(\mathbf{b}-\mathbf{a})=\frac{3}{5}\mathbf{a}+\frac{2}{5}\mathbf{b}\) with working shown
✓ 1
Part (b)
M1 \(\overrightarrow{OY}=10\overrightarrow{OX}\)
✓ 1
C1 the vectors are parallel and share the point O, so O, X, Y are collinear
✓ 1
Part (c)
B1 1 : 9
✓ 1
Exam tip: To prove three points are collinear, show two vectors between them are multiples of each other and state that they share a common point.
54IGME3-5-13r45-14Transformations described in wordsPaper HHard8 marks
Triangular glass panels in an atrium are drawn on a grid where 1 unit represents 1 m. Triangle ABC has A(1, 1), B(3, 1) and C(3, 2). It is enlarged with scale factor −2 and centre (1, −1) to give triangle A'B'C'.
Diagram on a coordinate grid
(a)
Find the coordinates of A', B' and C'.
(3)
(b)
Work out the area of triangle A'B'C'.
(2)
(c)
Describe fully the single transformation that maps triangle A'B'C' onto triangle ABC.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to enlarge from (1, −1) with a negative scale factor (image on the opposite side of the centre)
✓ 1
A1 two image points correct
✓ 1
A1 A'(1, −5), B'(−3, −5), C'(−3, −7)
✓ 1
Part (b)
M1 area of ABC = \(\frac{1}{2}\times 2\times 1\) = 1 and multiplied by \((-2)^2\) (or lengths 4 and 2)
✓ 1
A1 4 m2
✓ 1
Part (c)
B1 enlargement
✓ 1
B1 scale factor \(-\frac{1}{2}\)
✓ 1
B1 centre (1, −1)
✓ 1
Common error: A negative scale factor puts the image on the opposite side of the centre, and the area still scales by the square of the magnitude, which is always positive.
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