Printable · GCSE Foundation · ages 14-16
Translations as vectors worksheet — GCSE Foundation
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.
Translations as vectors worksheet — GCSE Foundation
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- 1.A robot on a grid moves by the vector , then by the vector , then by the vector . Write down the single column vector that has the same overall effect as these three moves.
- 2.Write down the column vector, from the options given, that would move a point to the right and downwards.
- 3.The point R(−6, 9) is translated by the vector . Write down the coordinates of the image of R.
- 4.In a phone game, a character starts at the point (−5, 2) on a grid. It moves by the vector 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.
- 5.The vector describes a translation. A student writes it as by mistake. Write down what error the student has made.
- 6.A drone starts at the point (2, −3) on a coordinate grid. It flies by the vector and then by the vector . Write down the column vector that would take the drone straight back to its starting point.
- 7.A shape is translated by the vector , and one vertex of the image is at the point (1, 6). The original shape is instead translated by the vector . Work out the coordinates of the image of that same vertex under this second translation.
- 8.The point M(3, 8) is translated by the vector . Write down the coordinates of the image of M.
- 9.A translation moves the point (1, 1) to the point (9, 4). Write down the column vector of this translation.
- 10.A point is translated twice by the vector . Write down the single column vector that describes the overall translation.
- 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.
- 12.A robotic arm's tip starts at the point (12.5, −4.25) on a grid measured in centimetres. It moves by the vector to pick up a component, then by the vector to place it. Work out the coordinates of the tip after both moves.
- 13.A student says the reverse of the translation vector is . Write down the correct reverse translation vector.
- 14.Vector A is and vector B is . Write down how vector B compares to vector A.
- 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 . Which of these points is NOT a vertex of the image?
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.
- (c) $\binom{4}{−3}$ — A positive top number moves a point to the right, and a negative bottom number moves it downwards, so $\binom{4}{−3}$ is right and down. $\binom{−4}{3}$ moves left and up — the opposite direction on both axes. $\binom{4}{3}$ moves right, like the key, but its positive bottom number moves it up, not down. $\binom{−4}{−3}$ moves down, like the key, but its negative top number moves it left, not right.
- (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.
- (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.
- (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 $\binom{-5}{-5}$.
- (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.
- (c) (−2, 10) — To translate M(3, 8) by $\binom{−5}{2}$, add −5 to the x-coordinate and 2 to the y-coordinate: (3 + (−5), 8 + 2) = (−2, 10). (5, 3) comes from swapping the two components of the vector before applying them. (−2, 8) applies only the x-component and forgets to add the y-component. (3, 10) applies only the y-component and forgets to add the x-component.
- (b) $\binom{8}{3}$ — The column vector is (end − start) in each coordinate: (9 − 1, 4 − 1) = (8, 3), written as $\binom{8}{3}$. $\binom{3}{8}$ swaps the horizontal and vertical components. $\binom{10}{5}$ comes from adding the coordinates instead of subtracting them. $\binom{−8}{−3}$ comes from working out start minus end instead of end minus start.
- (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.
- (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) (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.
- (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.
- (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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