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.The column vector describes a translation. Write down the movement it represents.
- 2.A student says the reverse of the translation vector is . Write down the correct reverse translation vector.
- 3.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.
- 4.Write down the column vector that describes a translation of 5 units to the left and 3 units down.
- 5.A translation is described by the column vector . Write down a description of this translation in words, stating the horizontal and vertical movement.
- 6.Vector A is and vector B is . Write down how vector B compares to vector A.
- 7.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 to a warehouse, then by the vector to the delivery address. Work out the coordinates of the delivery address.
- 8.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.
- 9.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.
- 10.The vector describes a translation. A student writes it as by mistake. Write down what error the student has made.
- 11.A translation is described by the column vector . Write down the column vector of the translation that undoes it.
- 12.The point A(2, 5) is translated to the point B(7, 3). Write down the column vector that describes this translation.
- 13.Write down the column vector that describes a translation of 4 units to the right and 4 units up.
- 14.A shape is translated by the vector . The point P(6, 1) lies on the shape. Write down the coordinates of the image of P.
- 15.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.
Answer key
- (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.
- (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.
- (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) $\binom{−5}{−3}$ — 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.
- (b) 7 units right and 2 units down — In the column vector $\binom{7}{−2}$, the top number gives the horizontal movement and the bottom number gives the vertical movement. A positive top number (7) means 7 units to the right, and a negative bottom number (−2) means 2 units down. The distractor '7 units left and 2 units up' reverses both signs, as if the vector were $\binom{−7}{2}$. The distractor '2 units right and 7 units down' swaps the horizontal and vertical numbers around. The distractor '7 units right and 2 units up' gets the horizontal movement correct but reverses the sign of the vertical movement.
- (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.
- (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) $\binom{0}{4}$ — 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, −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) 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.
- (c) $\binom{-6}{5}$ — Method: to undo a translation, travel the same journey backwards. A move of 6 to the right is undone by a move of 6 to the left, and a move of 5 down is undone by a move of 5 up, so both numbers change sign. Working: the top number 6 becomes −6 and the bottom number −5 becomes 5. Check by combining the two: 6 − 6 = 0 across and −5 + 5 = 0 up, so the shape finishes where it started. Answer: $\binom{-6}{5}$. Reversing only the horizontal movement gives $\binom{-6}{-5}$ and reversing only the vertical movement gives $\binom{6}{5}$, and each of those leaves the shape displaced. Swapping the two entries instead of changing their signs gives $\binom{-5}{6}$.
- (c) $\binom{5}{-2}$ — Method: the top number of a column vector is the change in the x-coordinate and the bottom number is the change in the y-coordinate, each worked out as image minus object. Working: across, 7 − 2 = 5; up, 3 − 5 = −2. So the point moves 5 to the right and 2 down. Answer: $\binom{5}{-2}$. Subtracting the other way round, object minus image, gives $\binom{-5}{2}$, which is the journey from B back to A. Recording the vertical change as 2 because the gap between 3 and 5 is 2, without noting the direction, gives $\binom{5}{2}$, a movement 2 upwards. Adding the coordinates instead of subtracting them gives $\binom{9}{8}$.
- (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.
- (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).
- (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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