Printable · GCSE Higher · ages 14-16
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.
part Higher
Transformations: rotation, reflection, translation, enlargement worksheet — GCSE Higher
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- 1.A pinhole camera is set up at a pitch-side stand, with its aperture at the point O = (2, 1) on a grid. Light from the corner F of a triangular flag, at the point (4, 2), passes through the aperture and forms an inverted image on the film behind it. The projection is an enlargement of scale factor −1.5 centred at O. Work out the coordinates of the image of corner F.
- 2.An enlargement with scale factor 5 maps the point (4, 1) to the point (12, 9). Which of these points is invariant (does not move) under this enlargement?
- 3.Triangle T has a vertex A at (4, 6). It is enlarged by a scale factor of −1/2, centre (2, 2). Work out the coordinates of the image of point A.
- 4.Postcard P is similar to postcard Q. The 6 cm side of P corresponds to the 15 cm side of Q, and the 8 cm side of P corresponds to a matching side of Q. Work out the length of that matching side of Q.
- 5.Point P(2, 5) is enlarged with centre (2, 1) to give image point P′(2, −7). Work out the scale factor of the enlargement.
- 6.Shape T has a vertex at (5, 3). T is enlarged by scale factor −1, centre (2, 1). Work out the coordinates of the image of the vertex (5, 3).
- 7.Triangle T has a vertex at (8, 4). It is enlarged by a scale factor of 1/2, centre (2, 4). Work out the coordinates of the image of this vertex.
- 8.Point E is at (4, 2). It is rotated 180° about the point (1, 1). Work out the coordinates of the image of point E.
- 9.Shape S has a vertex at (9, 6). It is enlarged by a scale factor of 1/3, centre the origin. Work out the coordinates of the image of this vertex.
- 10.Shape S is enlarged by a scale factor of −3 about a fixed centre. Which statement correctly describes the image compared to the original shape?
- 11.Triangle T has a vertex A at (3, 1). It is enlarged by a scale factor of −2, centre (1, 1). Work out the coordinates of the image of point A.
- 12.A postcard measuring 10 cm by 15 cm is enlarged by a scale factor of 2 to make a poster. Work out the area of the poster, in cm².
- 13.Shape S is enlarged by a scale factor of 3, centre the origin. A vertex of S is at (2, 1). Work out the coordinates of the image of this vertex.
- 14.Point D is at (3, 7). It is reflected in the line y = x. Work out the coordinates of the image of point D.y = x
- 15.Triangle ABC is similar to triangle PQR. Side AB is 8 cm and corresponds to side PQ, which is 12 cm. Side AC is 6 cm and corresponds to side PR. Work out the length of PR.
Answer key
- (a) (−1, −0.5) — Method: for an enlargement centred at a point O that is not the origin, first find the point's position relative to O, multiply that by the scale factor, then add O's coordinates back on — image = O + k × (F − O). Working: F relative to O is (4, 2) − (2, 1) = (2, 1). Multiplying by the scale factor, −1.5 × (2, 1) = (−3, −1.5). Adding O back on: (2, 1) + (−3, −1.5) = (−1, −0.5). Answer: (−1, −0.5). Measure F's position from the aperture O, not from the origin, apply the full signed scale factor −1.5 to that displacement, and add O's own coordinates back on afterwards: measuring from the origin instead of O, treating the factor as +1.5, or adding O onto the wrong displacement all place the image mark somewhere else.
- (d) (2, −1) — Method: for an enlargement, image = centre + k × (point − centre), so the centre satisfies centre = (image − k × point) ÷ (1 − k). Working: with k = 5, point (4, 1) and image (12, 9): 5 × (4, 1) = (20, 5); (12, 9) − (20, 5) = (−8, 4); dividing by 1 − 5 = −4 gives (2, −1). Answer: (2, −1), the centre of the enlargement, is the only invariant point since the scale factor is not 1. Subtracting the point itself instead of k times the point, (12, 9) − (4, 1) = (8, 8), then dividing by −4 gives (−2, −2); dividing by k − 1 = 4 instead of 1 − k = −4 gives (−2, 1); and simply taking the midpoint of the point and its image ignores the scale factor altogether and gives (8, 5). The centre of an enlargement is never just the midpoint between a point and its image unless the scale factor happens to be −1 — always use the full centre formula and keep the scale factor k in it.
- (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) 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.
- (d) −2 — Method: the scale factor is the ratio of the image vector to the object vector, both measured FROM THE CENTRE of enlargement, keeping every sign. Working: the vector from the centre (2, 1) to P(2, 5) is (0, 4); the vector from the centre to P′(2, −7) is (0, −8). The scale factor is −8 ÷ 4 = −2. Answer: −2. Measure both vectors from the CENTRE, not from the origin, divide the IMAGE vector by the OBJECT vector and not the other way round, and keep the negative sign: a negative scale factor is not the same size as its positive counterpart with the sign dropped.
- (b) (−1, −1) — An enlargement by scale factor −1, centre (2, 1), sends a point P to the point on the opposite side of the centre, the same distance away: the image is 2 × centre − P. For the vertex (5, 3), this gives (2 × 2 − 5, 2 × 1 − 3) = (4 − 5, 2 − 3) = (−1, −1). Treating the centre as though it were the origin, and simply negating the point's coordinates, gives (−5, −3) — this ignores that the true centre is (2, 1), not (0, 0). Using scale factor +1 instead of −1 leaves the point exactly where it started, at (5, 3). Adding the point's displacement from the centre instead of subtracting it gives (2 × 2 + 5, 2 × 1 + 3) = (9, 5). Double the centre and subtract the point, and the image is (−1, −1).
- (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).
- (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).
- (c) (3, 2) — For an enlargement centred on the origin, multiply every coordinate by the scale factor: (9 × 1/3, 6 × 1/3) = (3, 2). A pupil who multiplies by 3 instead of by 1/3 gets (27, 18). A pupil who subtracts a third of each coordinate instead of scaling by a third gets (9 − 3, 6 − 2) = (6, 4). A pupil who applies the scale factor to the x-coordinate only gets (3, 6). The correct image is (3, 2).
- (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.
- (c) (−3, 1) — Method: for an enlargement about a centre, first find the vector from the centre to the point, multiply it by the scale factor, INCLUDING its sign, then add the result back onto the centre. Working: the vector from the centre (1, 1) to A(3, 1) is (3 − 1, 1 − 1) = (2, 0). Multiplying by the scale factor −2 gives −2 × 2 = −4 and −2 × 0 = 0, so the scaled vector is (−4, 0). Adding this to the centre gives 1 + (−4) = −3 and 1 + 0 = 1, so the image is (−3, 1). Answer: (−3, 1). A NEGATIVE scale factor keeps its sign all the way through the calculation: do not treat −2 as +2, and do not treat it as a fraction like 1/2, which is the rule for a scale factor between 0 and 1, not a negative one. Always measure the vector from the CENTRE of enlargement, never from the origin, unless the two happen to coincide.
- (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.
- (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) (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).
- (c) 9 cm — The scale factor from triangle ABC to triangle PQR is PQ ÷ AB = 12 ÷ 8 = 1.5. So PR = AC × 1.5 = 6 × 1.5 = 9 cm. (4 cm comes from using the scale factor upside down, AB ÷ PQ; 10 cm comes from adding the difference between AB and PQ, 4 cm, to AC instead of scaling; 16 cm comes from multiplying PQ by the ratio AB : AC instead of scaling AC by the correct scale factor.)
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