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Translations as vectors worksheet — GCSE Higher

Fifteen questions on "translations as vectors" — DfE statement G24. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.

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Translations as vectors worksheet — GCSE Higher

Maths · GCSE Higher · G24 · Translations as vectors · 15 questions

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1.A courier drone delivers a parcel. It starts at the point (2, 1) on a map grid measured in kilometres. It flies by the vector $(- 3 5)$ to a warehouse, then by the vector $(4 - 1)$ to the delivery address. Work out the coordinates of the delivery address.
(a)(7, −2)
(b)(11, −3)
(c)(8, −6)
(d)(6, 2)
2.A shape is translated by the vector $(- 7 4)$ and then by the vector $(3 - 9)$ . Write down the single column vector that has the same effect as those two translations together.
(a) $(- 4 - 5)$
(b) $(- 10 13)$
(c) $(- 4 - 13)$
(d) $(- 10 - 5)$
3.A robot on a grid moves by the vector $(- 2 3)$ , then by the vector $(5 - 7)$ , then by the vector $(1 4)$ . Write down the single column vector that has the same overall effect as these three moves.
(a) $(- 6 14)$
(b) $(0 4)$
(c) $(3 - 4)$
(d) $(4 0)$
4.A drone starts at the point (3.5, −2) on a grid measured in metres. It flies by the vector $(4.5 - 6.2)$ to check a first sensor. The drone then needs to fly in a straight line to reach the point (−1.7, 9) to check a second sensor. Work out the column vector of this second flight.
(a) $(6.5 1)$
(b) $(- 6.5 - 1)$
(c) $(6.5 - 4.4)$
(d) $(11 - 5.2)$
5.Rectangle ABCD has vertices A(−1, 2), B(4, 2), C(4, 5) and D(−1, 5). The rectangle is translated so that C moves to (0, −1). Write down the coordinates of the image of A.
(a)(3, 8)
(b)(−5, 8)
(c)(0, −1)
(d)(−5, −4)
6.A robotic arm's tip starts at the point (12.5, −4.25) on a grid measured in centimetres. It moves by the vector $(6.5 - 3.75)$ to pick up a component, then by the vector $(- 1.5 2.25)$ to place it. Work out the coordinates of the tip after both moves.
(a)(0.75, 11)
(b)(8.75, 2.25)
(c)(6.5, 3.75)
(d)(11, 0.75)
7.A student says the reverse of the translation vector $(- 9 4)$ is $(9 4)$ . Write down the correct reverse translation vector.
(a) $(- 9 - 4)$
(b) $(- 9 4)$
(c) $(9 - 4)$
(d) $(9 4)$
8.After a translation by the vector $(6 - 2)$ , a point lands on (4, 1). Write down the coordinates of the point before the translation.
(a)(2, −5)
(b)(2, 7)
(c)(6, −5)
(d)(6, 7)
9.Vertex X of a triangle is at (−3, 5). After a translation, the image of X is at (2, −1). Write down the column vector of this translation.
(a) $(- 6 5)$
(b) $(6 5)$
(c) $(6 - 5)$
(d) $(- 6 - 5)$
10.The point R(−6, 9) is translated by the vector $(- 11 5)$ . Write down the coordinates of the image of R.
(a)(−1, 9)
(b)(−1, −2)
(c)(−6, −2)
(d)(−11, 20)
11.A ship's radar shows a lighthouse at the point (12, −4) on a grid measured in nautical miles. The ship is at (2, 5). The ship sails along the vector that takes it directly to the lighthouse, then sails along that same vector again. Work out the ship's final position.
(a)(−18, 23)
(b)(22, −13)
(c)(2, 5)
(d)(12, −4)
12.The vector $(- 2 5)$ describes a translation. A student writes it as $(5 - 2)$ by mistake. Write down what error the student has made.
(a)Doubled the x component
(b)Reversed both signs
(c)Swapped the x and y components
(d)Reversed only the y sign
13.Vector A is $(- 5 2)$ and vector B is $(5 - 2)$ . Write down how vector B compares to vector A.
(a)B is the same as A
(b)B is unrelated to A in direction
(c)B is twice A
(d)B is the reverse of A
14.A point is translated twice by the vector $(5 - 3)$ . Write down the single column vector that describes the overall translation.
(a) $(5 - 3)$
(b) $(10 - 6)$
(c) $(5 - 6)$
(d) $(15 - 9)$
15.Quadrilateral ABCD has vertices A(1, 1), B(5, 1), C(5, 4) and D(1, 4). The quadrilateral is translated by the vector $(3 - 2)$ . Which of these points is NOT a vertex of the image?
(a)(−1, 1)
(b)(3, 4)
(c)(−1, 4)
(d)(3, 7)

Answer key

(d) (6, 2) — Applying the first vector: (2, 1) + (5, −3) = (7, −2), which is the warehouse. Applying the second vector: (7, −2) + (−1, 4) = (6, 2), the delivery address. '(7, −2)' stops at the warehouse and forgets the second flight. '(8, −6)' comes from adding (1, −4) instead of (−1, 4) for the second vector, getting both signs wrong. '(11, −3)' comes from swapping the components of the second vector to (4, −1) before adding.
(a) $\binom{-5}{-4}$ — Method: one translation followed by another is a single translation, and the two vectors are added: top to top, bottom to bottom. Working: across, 4 − 9 = −5; up, −7 + 3 = −4. Answer: $\binom{-5}{-4}$. Subtracting the second vector instead of adding it gives 13 on top and −7 − 3 = −10 underneath. Adding the top numbers correctly but subtracting the bottom ones gives −10 underneath with −5 on top. Adding 4 and 9 as though both were positive and then keeping the minus sign of the larger gives −13 on top.
(d) $\binom{0}{4}$ — Add the three vectors component by component: x: 3 + (−7) + 4 = 0; y: −2 + 5 + 1 = 4, giving $\binom{0}{4}$. $\binom{−4}{3}$ comes from adding only the first two vectors and forgetting the third. $\binom{14}{−6}$ comes from reading the second vector as $\binom{7}{−5}$ instead of $\binom{−7}{5}$, flipping its signs. $\binom{4}{0}$ comes from swapping the final x-total and y-total.
(a) $\binom{1}{6.5}$ — After the first flight, the drone is at (3.5 − 6.2, −2 + 4.5) = (−2.7, 2.5). The second vector takes it from (−2.7, 2.5) to (−1.7, 9): subtract the coordinates, (−1.7 − (−2.7), 9 − 2.5) = (1, 6.5). The distractor $\binom{−4.4}{6.5}$ comes from treating the drone's position after the first flight as (2.7, 2.5) instead of (−2.7, 2.5), giving −1.7 − 2.7 = −4.4 for the top number. The distractor $\binom{−5.2}{11}$ comes from finding the vector straight from the start point (3.5, −2) to (−1.7, 9), ignoring the first flight altogether. The distractor $\binom{−1}{−6.5}$ comes from subtracting the wrong way round, (−2.7 − (−1.7), 2.5 − 9), which reverses both signs of the correct vector.
(d) (−5, −4) — Method: every vertex of a translated shape moves by the same vector, so find that vector from the one vertex whose image is given, then apply it to A. Working: C(4, 5) moves to (0, −1), so across 0 − 4 = −4 and up −1 − 5 = −6, giving the vector $\binom{-4}{-6}$. Applying it to A(−1, 2): −1 − 4 = −5 and 2 − 6 = −4. Answer: the image of A is (−5, −4). Working the vector out as object minus image gives 4 to the right and 6 up, which applied to A gives (3, 8). Getting the horizontal movement right but reversing the vertical one gives (−5, 8). Treating (0, −1) as the image of every vertex ignores that a translation carries each vertex to a different place.
(d) (11, 0.75) — Add the moves to the starting point one component at a time. x: 12.5 + (−3.75) + 2.25 = 11; y: −4.25 + 6.5 + (−1.5) = 0.75, giving (11, 0.75). (8.75, 2.25) stops after the first move only and never applies the second vector. (6.5, 3.75) comes from subtracting the second vector instead of adding it. (0.75, 11) comes from swapping the final x-coordinate and y-coordinate.
(c) $\binom{-4}{9}$ — The reverse of a translation negates both components, so the reverse of $\binom{4}{-9}$ is $\binom{-4}{9}$. $\binom{4}{9}$ is the student's mistake — only the bottom number has been negated, and the top number was left unchanged. $\binom{-4}{-9}$ makes the opposite error, negating only the top number. $\binom{4}{-9}$ is simply the original vector, unchanged.
(c) (6, −5) — Method: the translation has already happened, so it must be undone: reverse the vector and apply the reverse to the point that is given. Working: reversing $\binom{-2}{6}$ gives $\binom{2}{-6}$, so the x-coordinate is 4 + 2 = 6 and the y-coordinate is 1 − 6 = −5. Check: from (6, −5) the given vector gives 6 − 2 = 4 and −5 + 6 = 1, which is the point named in the question. Answer: (6, −5). Applying the vector forwards instead of backwards gives (2, 7). Reversing the horizontal movement but not the vertical one gives (6, 7), and reversing the vertical movement but not the horizontal one gives (2, −5).
(a) $\binom{5}{−6}$ — The vector is (image − original) in each coordinate: (2 − (−3), −1 − 5) = (5, −6). $\binom{−5}{6}$ comes from working out original − image instead of image − original. $\binom{5}{6}$ gets the x-component right but makes a sign error on the y-component. $\binom{−5}{−6}$ gets the y-component right but makes a sign error on the x-component, working out −3 − 2 = −5 instead of 2 − (−3) = 5.
(b) (−1, −2) — To translate R(−6, 9) by $\binom{5}{−11}$, add 5 to the x-coordinate and −11 to the y-coordinate: (−6 + 5, 9 + (−11)) = (−1, −2). (−1, 9) applies only the x-component and leaves the y-coordinate unchanged. (−6, −2) applies only the y-component and leaves the x-coordinate unchanged. (−11, 20) comes from subtracting the vector instead of adding it.
(b) (22, −13) — The vector from the ship to the lighthouse is (12, −4) − (2, 5) = (10, −9). Sailing along this vector twice from the start gives (2, 5) + 2 × (10, −9) = (2 + 20, 5 − 18) = (22, −13). '(12, −4)' stops after the ship reaches the lighthouse and ignores the second identical leg. '(−18, 23)' comes from finding the vector the wrong way round, as (2, 5) − (12, −4) = (−10, 9), and then doubling that. '(2, 5)' comes from adding the vector and then subtracting it again, wrongly cancelling the two legs instead of adding them.
(c) Swapped the x and y components — The student's vector has the same two numbers, 5 and −2, but in swapped positions, so the error is swapping the x and y components rather than an error with signs or size. 'Reversed both signs' is wrong because the numbers 5 and −2 have not changed sign, only position. 'Reversed only the y sign' is wrong for the same reason — no sign has actually changed. 'Doubled the x component' is wrong because neither number has changed in size.
(d) B is the reverse of A — Every component of vector B is the negative of the matching component of vector A (−2 is the negative of 2, and 5 is the negative of −5), so B undoes the translation that A performs — B is the reverse of A. 'B is the same as A' ignores that both signs have flipped. 'B is twice A' confuses a sign change with a scale-factor change; the sizes of the components have not changed, only their signs. 'B is unrelated to A in direction' misses that the two vectors are directly linked, just in opposite directions.
(b) $\binom{-6}{10}$ — Translating twice by the same vector doubles both components: 2 × $\binom{-3}{5}$ = $\binom{-6}{10}$. $\binom{-3}{5}$ forgets to double the vector at all, giving only one translation's worth. $\binom{-9}{15}$ trebles the vector instead of doubling it. $\binom{-6}{5}$ doubles only the top number and forgets to double the bottom number.
(a) (−1, 1) — Translating by $\binom{−2}{3}$ subtracts 2 from every x-coordinate and adds 3 to every y-coordinate. This gives image vertices (−1, 4), (3, 4), (3, 7) and (−1, 7). The point (−1, 1) is not one of these: it has the correct new x-coordinate (1 − 2 = −1) but keeps the original y-coordinate (1) instead of adding 3, as if only the horizontal part of the vector had been applied.

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