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Transformations: rotation, reflection, translation, enlargement worksheet — GCSE Higher

Fifteen questions on "transformations: rotation, reflection, translation, enlargement" — DfE statement G7. 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: Transformations: rotation, reflection, translation, enlargement worksheet — GCSE Higher

Maths · GCSE Higher · G7 · Transformations: rotation, reflection, translation, enlargement · 15 questions

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(d) 20 cm — The scale factor from P to Q is 15 ÷ 6 = 2.5. Apply the same scale factor to the other side: 8 × 2.5 = 20 cm. A pupil who divides instead of multiplying by the scale factor gets 8 ÷ 2.5 = 3.2 cm. A pupil who adds the difference between the two known sides, 15 − 6 = 9, to 8 instead of scaling gets 8 + 9 = 17 cm. A pupil who rounds the scale factor 2.5 down to 2 gets 8 × 2 = 16 cm. The correct length is 20 cm.
(c) (1, 0) — To enlarge about a centre other than the origin, find the vector from the centre to the point, scale that vector, then add it back to the centre. The vector from (2, 2) to A(4, 6) is (2, 4). Scaling by −1/2 gives (−1, −2). Adding this to the centre (2, 2) gives the image point (1, 0). (3, 4) comes from using +1/2 instead of −1/2, so the image lands on the same side as A instead of the opposite side. (−2, −3) comes from scaling A's coordinates directly about the origin, ignoring that the centre is (2, 2). (−2, −6) comes from using a scale factor of −2 instead of −1/2.
(d) 600 cm² — Enlarging by scale factor 2 makes the new dimensions 10 × 2 = 20 cm and 15 × 2 = 30 cm, so the poster's area = 20 × 30 = 600 cm². A pupil who scales the original area, 150 cm², by the scale factor itself instead of by its square gets 150 × 2 = 300 cm². A pupil who adds the scale factor to each dimension instead of multiplying gets (10 + 2) × (15 + 2) = 204 cm². A pupil who forgets to enlarge the postcard at all just uses the original area, 150 cm². The correct area of the poster is 600 cm².
(d) (−2, 0) — A 180° rotation about a centre (a, b) maps (x, y) to (2a − x, 2b − y). Here that gives (2 × 1 − 4, 2 × 1 − 2) = (−2, 0). A pupil who rotates about the origin instead of (1, 1) gets (−4, −2). A pupil who adds the centre's coordinates instead of applying the rotation formula gets (4 + 1, 2 + 1) = (5, 3). A pupil who just subtracts the centre's coordinates from E's, without doubling and reversing, gets (4 − 1, 2 − 1) = (3, 1). The correct image is (−2, 0).
(a) 3 times the size, opposite side, rotated 180° — Method: for any enlargement, the MAGNITUDE of the scale factor gives the size ratio between image and object, while the SIGN decides which side of the centre the image falls on; a negative scale factor puts the image on the opposite side, which is equivalent to a 180° rotation about the centre. Working: the scale factor is −3, so the size ratio is the magnitude, which is 3, and the negative sign puts the image on the opposite side of the centre, rotated 180° relative to the original. Answer: 3 times the size, opposite side, rotated 180°. The magnitude of the scale factor controls the SIZE only: do not let the sign leak into it and turn 3 into 1/3. The sign controls the SIDE and orientation, which a positive-only view of enlargement, just 'further away' with the same orientation, misses entirely.
(b) (6, 3) — For an enlargement centred on the origin, multiply both coordinates by the scale factor: (2 × 3, 1 × 3) = (6, 3). A pupil who adds the scale factor to each coordinate instead of multiplying gets (2 + 3, 1 + 3) = (5, 4). A pupil who multiplies only the x-coordinate gets (6, 1). A pupil who multiplies only the y-coordinate gets (2, 3). The correct image is (6, 3).
(b) 45 cm² — Area scales with the square of the linear scale factor, and squaring a negative number gives a positive result: (−3)² = 9. The area of T is 5 × 9 = 45 cm². 15 cm² comes from multiplying the original area by the scale factor directly (5 × 3), without squaring. 9 cm² is the area scale factor itself, (−3)², with the multiplication by the original area 5 cm² left out. −15 cm² comes from multiplying 5 × (−3) and carrying the negative sign through, without squaring at all.
(a) No — it's a 180° rotation, not a reflection. — A negative scale factor scales the shape by the size of the factor and turns it through 180° about the centre of enlargement — every point of the image lands on the opposite side of the centre from the object. A 180° turn is a rotation, and a rotation never reverses a shape's handedness (its sense), whereas a reflection always does, so the image is not a mirror image: Priti is wrong. The option claiming negative numbers always mean a reflection confuses the sign of a scale factor with the transformation of reflection — they are not the same thing. The option saying only the size changes ignores the sign altogether: the 180° turn does change the way the shape faces, even though it leaves the handedness unchanged. The option equating 'upside down' with 'mirrored' assumes the two always mean the same thing, but a 180° rotation and a reflection in a horizontal line can both look 'upside down' while being different transformations — only the reflection reverses the shape's handedness.
(a) (5, 4) — Method: find the vector from the centre to the point, multiply it by the scale factor, then add the result back to the centre. Working: the vector from (2, 4) to (8, 4) is (6, 0); multiplying by 1/2 gives (3, 0); adding this to the centre (2, 4) gives (5, 4). Options: (4, 2) comes from multiplying the original coordinates by 1/2 directly, ignoring the centre of enlargement; (14, 4) comes from using a scale factor of 2 instead of 1/2, giving (2, 4) + 2×(6, 0) = (14, 4); (8, 2) comes from halving only the y-coordinate and leaving the x-coordinate unchanged. Answer: (5, 4).
(a) 3 cm² — Area scale factor = (linear scale factor)² = (1/4)² = 1/16. Area of T = 48 × 1/16 = 3 cm². (12 cm² comes from multiplying by the linear scale factor 1/4 directly, without squaring it; 24 cm² comes from taking the square root of the scale factor instead of squaring it; 768 cm² comes from squaring the reciprocal of the scale factor, 4, instead of the scale factor itself.)
(c) A rotation of 180° about the origin — An enlargement by scale factor −1 sends every point (x, y) to (−x, −y) — both coordinates change sign. A rotation of 180° about the origin does exactly the same thing to every point, so the two transformations have identical effect. A reflection in the x-axis only changes the sign of the y-coordinate, sending (x, y) to (x, −y), leaving the x-coordinate untouched. A reflection in the y-axis only changes the sign of the x-coordinate, sending (x, y) to (−x, y), leaving the y-coordinate untouched. Treating a negative scale factor as though it behaves like a positive one gives no transformation at all, but the minus sign is not decorative — it reverses both coordinates. Both signs flip together, which is exactly what a 180° rotation about the origin does.
(a) −2 — Let the centre be C = (c, 0). The vector from the centre to the image equals the scale factor k times the vector from the centre to the object: (7 − c, −4) = k(1 − c, 2). The y-coordinate gives −4 = 2k, so k = −2 — this doesn't depend on knowing c. (Checking: substituting k = −2 into the x-equation gives c = 3, consistent with a centre on the x-axis.) '2' comes from taking the magnitude of the ratio without noticing the image is on the opposite side of the centre from the object, so the sign should be negative. '−1/2' comes from inverting the scale factor, dividing the object's coordinate by the image's instead of the other way round. '3' is the x-coordinate of the centre, mistaken for the scale factor.
(b) 4 — The scale factor is the distance from the centre to the image, divided by the distance from the centre to the object: (5 − 1) ÷ (2 − 1) = 4 ÷ 1 = 4. (2.5 comes from dividing the raw y-coordinates, 5 ÷ 2, without first subtracting the centre's coordinate; 3 comes from subtracting the two distances instead of dividing them; 0.25 comes from dividing the distances the wrong way round.)
(d) £48 — Scale factor = new width ÷ original width = 40 ÷ 10 = 4. Poster height = 15 × 4 = 60 cm. Cost = 60 × £0.80 = £48. (£12 comes from forgetting to scale the height at all, and pricing the original 15 cm height; £15.20 comes from adding the scale factor 4 to the height instead of multiplying by it; £3 comes from dividing the height by the scale factor instead of multiplying by it.)
(b) (7, 3) — Reflecting in the line y = x swaps the x- and y-coordinates: (3, 7) → (7, 3). A pupil who reflects in the x-axis instead gets (3, −7). A pupil who reflects in the y-axis instead gets (−3, 7). A pupil who confuses y = x with y = −x, swapping the coordinates and changing both signs, gets (−7, −3). The correct image is (7, 3).

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