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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 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) $(4 0)$
2.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
3.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) $(15 - 9)$
4.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) $(- 6 - 5)$
5.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)
6.A shape is translated by the vector $(3 - 8)$ , and one vertex of the image is at the point (1, 6). The original shape is instead translated by the vector $(- 5 2)$ . Work out the coordinates of the image of that same vertex under this second translation.
(a)(7, 8)
(b)(11, 3)
(c)(3, 1)
(d)(11, −2)
7.A crane's hook starts at the point (3.4, 9.5) on a construction site plan measured in metres. It moves along the vector $(- 4.5 - 1.2)$ to pick up a beam, then along the vector $(- 2.5 0.8)$ to place it. Work out the single column vector that describes the hook's overall movement from its start position to where it places the beam.
(a) $(- 2 - 2)$
(b) $(- 7 - 0.4)$
(c) $(7 - 0.4)$
(d) $(- 7 2)$
8.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)
9.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)
10.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)
11.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)
12.A drone starts at the point (2, −3) on a coordinate grid. It flies by the vector $(4 6)$ and then by the vector $(- 9 - 1)$ . Write down the column vector that would take the drone straight back to its starting point.
(a) $(- 5 - 5)$
(b) $(5 - 5)$
(c) $(- 5 5)$
(d) $(- 13 - 7)$
13.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)
14.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) $(- 10 - 5)$
15.In a phone game, a character starts at the point (−5, 2) on a grid. It moves by the vector $(- 9 6)$ to collect a coin. The player now wants the character's next single move to finish at the point (3, 0). Write down the column vector of that second move.
(a) $(- 2 8)$
(b) $(- 7 - 2)$
(c) $(7 2)$
(d) $(- 7 4)$

Answer key

(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.
(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) $\binom{5}{−6}$ — 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.
(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.
(d) (11, −2) — First undo the original translation to find the vertex on the original shape: (1 − (−8), 6 − 3) = (9, 3). Then apply the second vector to that original vertex: (9 + 2, 3 + (−5)) = (11, −2). (3, 1) comes from applying the second vector to the image point (1, 6) instead of to the original vertex — (1 + 2, 6 + (−5)) = (3, 1). (7, 8) comes from subtracting the second vector from the original vertex (9, 3) instead of adding it — (9 − 2, 3 − (−5)) = (7, 8). (11, 3) comes from applying only the x-component of the second vector to the original vertex and leaving the y-coordinate unchanged.
(b) $\binom{-0.4}{-7}$ — The overall movement is the sum of the two vectors: (−1.2 + 0.8, −4.5 + (−2.5)) = (−0.4, −7). '$\binom{-2}{-2}$' comes from subtracting the second vector from the first instead of adding them. '$\binom{-0.4}{7}$' gets the top number right but drops the negative sign on the bottom number. '$\binom{2}{-7}$' comes from treating −1.2 + 0.8 as if the signs did not matter, giving +2 instead of −0.4.
(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.
(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) (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.
(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) $\binom{-5}{5}$ — Method: add the two flights to get the single vector of the whole journey out, then reverse that vector to get the journey home. Working: across, 6 − 1 = 5; up, 4 − 9 = −5. So the drone finishes at (7, −8), which is 5 to the right of its start and 5 below it. The way home is therefore 5 to the left and 5 up. Answer: $\binom{-5}{5}$. Giving the combined journey itself, $\binom{5}{-5}$, describes the flight out rather than the flight home. Subtracting the second vector instead of adding it makes the journey out 7 across and 13 up, and reversing that gives $\binom{-7}{-13}$. Reversing the horizontal movement but leaving the vertical one alone gives $\binom{-5}{-5}$.
(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) $\binom{-5}{-4}$ — 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.
(c) $\binom{2}{7}$ — First find the character's position after the first move: (−5 + 6, 2 + (−9)) = (1, −7). The second move takes it from (1, −7) to (3, 0), so subtract the current position from the target: (3 − 1, 0 − (−7)) = $\binom{2}{7}$. $\binom{8}{−2}$ works from the starting point (−5, 2) and ignores the first move — (3 − (−5), 0 − 2) = $\binom{8}{−2}$. $\binom{−2}{−7}$ subtracts the wrong way round, current position minus target, instead of target minus current position — (1 − 3, −7 − 0) = $\binom{−2}{−7}$. $\binom{4}{−7}$ adds the target's coordinates to the current position instead of subtracting — (1 + 3, −7 + 0) = $\binom{4}{−7}$.

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