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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 · Non-calculator · 15 questions

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1.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) $(- 13 - 7)$
2.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)
3.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)
4.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) $(9 4)$
5.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)
6.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) $(4 0)$
7.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)
8.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) $(- 10 - 5)$
9.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) $(- 7 4)$
10.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
11.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)
12.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)
13.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) $(- 6 - 5)$
14.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)
15.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) $(- 7 2)$

Answer key

(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 $\binom{-5}{-5}$.
(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.
(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.
(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.
(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) $\binom{0}{4}$ — 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) (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.
(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) = $\binom{4}{−7}$.
(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.
(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.
(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.
(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) $\binom{-0.4}{-7}$ — 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.

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