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.
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Answer key: Translations as vectors worksheet — GCSE Foundation
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- (b) (3, 5) — Method: add the top number of the vector to the x-coordinate and the bottom number to the y-coordinate. Working: adding −3 to the x-coordinate is 6 − 3 = 3, and adding 4 to the y-coordinate is 1 + 4 = 5. Answer: the image of P is (3, 5). Subtracting the vector instead of adding it reverses the translation and gives (9, −3). Adding the top number but subtracting the bottom one, on the assumption that the lower entry always means downwards, gives (3, −3); the minus sign in a vector has already recorded the direction. Reading the two numbers the wrong way round, so that the shape moves 4 across and 3 down, gives (10, −2).
- (a) 6 left, 2 up — In a column vector $\binom{x}{y}$, the top number gives the horizontal movement (negative means left) and the bottom number gives the vertical movement (positive means up), so $\binom{-6}{2}$ means 6 units left and 2 units up. '6 right, 2 up' ignores the negative sign on the top number. '2 left, 6 up' swaps the top and bottom numbers. '6 left, 2 down' ignores the positive sign on the bottom number.
- (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.
- (d) $\binom{−5}{−3}$ — Left is a negative horizontal move and down is a negative vertical move, so the vector is $\binom{−5}{−3}$. $\binom{5}{−3}$ comes from treating 'left' as a positive move. $\binom{−5}{3}$ comes from treating 'down' as a positive move. $\binom{−3}{−5}$ comes from swapping the horizontal and vertical components.
- (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) $\binom{1}{6.5}$ — 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.
- (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) $\binom{-4}{9}$ — 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) $\binom{-0.4}{-7}$ — 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.
- (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) (−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.
- (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.
- (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) (−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.
- (a) $\binom{4}{4}$ — A movement to the right is a positive top number and a movement up is a positive bottom number, so 4 right and 4 up gives $\binom{4}{4}$. $\binom{4}{-4}$ wrongly gives the vertical movement a negative sign, as if it were downward. $\binom{-4}{4}$ wrongly gives the horizontal movement a negative sign, as if it were leftward. $\binom{-4}{-4}$ gets both signs wrong, as if the translation were left and down.
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