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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Translations as vectors worksheet — GCSE Foundation
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- 1.The point R(−6, 9) is translated by the vector . Write down the coordinates of the image of R.
- 2.A pin on a map is at grid reference (5, 2). It is moved by the vector . Write down its new grid reference.
- 3.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.
- 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.Write down the column vector that describes a translation of 4 units to the right and 4 units up.
- 6.After a translation by the vector , a point lands on (4, 1). Write down the coordinates of the point before the translation.
- 7.The point A(2, 5) is translated to the point B(7, 3). Write down the column vector that describes this translation.
- 8.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.
- 9.The point M(3, 8) is translated by the vector . Write down the coordinates of the image of M.
- 10.A translation is described by the column vector . Write down a description of this translation in words, stating the horizontal and vertical movement.
- 11.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.
- 12.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?
- 13.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.
- 14.A shape is translated by the vector and then by the vector . Write down the single column vector that has the same effect as those two translations together.
- 15.Vector A is and vector B is . Write down how vector B compares to vector A.
Answer key
- (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) (2, 6) — Adding the vector to the point gives the new position: (5 + (−3), 2 + 4) = (2, 6). '(8, 6)' comes from adding 3 instead of subtracting it, treating the top number as positive. '(9, −1)' comes from swapping the two components of the vector before adding them. '(2, 2)' applies the horizontal movement correctly but forgets to add the vertical movement.
- (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) $\binom{2}{7}$ — 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) = $\binom{4}{−7}$.
- (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.
- (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).
- (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 $\binom{9}{8}$.
- (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.
- (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) 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.
- (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) (−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.
- (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.
- (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.
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