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.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.
- 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 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.
- 4.Point B is at (3, 5). It is reflected in the line y = 2. Work out the coordinates of the image of point B.
- 5.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.
- 6.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.
- 7.Triangle A has a vertex at (1, 2). Triangle A is enlarged, centre (1, 1), to give triangle B, whose corresponding vertex is at (1, 5). Work out the scale factor of the enlargement.
- 8.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.
- 9.Triangle J is mapped onto triangle K by a single transformation with centre O. Triangle K is half the size of triangle J and is inverted (turned through 180°). Which of these transformations maps J onto K?
- 10.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.
- 11.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?
- 12.A scale model of a bridge is built so that 2 cm on the model represents 5 m of the real bridge. The real bridge is 45 m long. Work out the length of the model bridge, in centimetres.
- 13.Triangle T has a vertex at (2, 1). It is enlarged by a scale factor of 3, centre (0, 0). 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.Shape S is enlarged by scale factor −1, centre (0, 0). Which single transformation has the same overall effect as this enlargement, for every point?
Answer key
- (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).
- (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.
- (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).
- (b) (3, −1) — Method: find the distance from the point to the mirror line, then place the image the same distance on the other side of the line. Working: B is 5 − 2 = 3 units above the line y = 2, so its image is 3 units below the line, at y = 2 − 3 = −1, giving (3, −1); the x-coordinate does not change, since the mirror line is horizontal. Options: (3, −5) comes from reflecting in the x-axis (y = 0) instead of the line y = 2; (−3, 5) comes from reflecting in the y-axis instead of a horizontal line; (3, −4) comes from doubling the distance from the line instead of reflecting it, giving 2 − 2×3 = −4. Answer: (3, −1).
- (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) 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.
- (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) −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) An enlargement, centre O, scale factor −1/2. — A single enlargement with a negative scale factor both changes the size (by the magnitude of the factor) and rotates the image 180° about the centre (giving the inverted orientation) in one transformation. Halving the size needs a magnitude of 1/2, and inverting needs a negative sign, so the scale factor is −1/2, centre O. 'Scale factor 1/2' gives the correct size but no inversion, since a positive scale factor keeps the same orientation. 'A 180° rotation with no change in size' gives the inversion but not the halving — K is described as smaller than J, so size must change too. 'Scale factor −2' inverts correctly but doubles the size instead of halving it.
- (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.
- (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) 18 — Method: divide the real length by 5 to find how many 'units' of 5 m it contains, then multiply by 2 cm for each unit. Working: 45 ÷ 5 = 9, so the real bridge is 9 lots of 5 m; each lot is represented by 2 cm on the model, so the model length is 9 × 2 = 18 cm. Options: 9 comes from stopping after the division, without multiplying by the 2 cm per unit; 90 comes from multiplying the real length by 2 directly, without dividing by 5 first; 4.5 comes from dividing by 5 and then dividing by 2 again, instead of multiplying by 2. Answer: 18.
- (d) (6, 3) — For an enlargement centred at the origin, each coordinate is multiplied by the scale factor: (2, 1) → (2 × 3, 1 × 3) = (6, 3). ((5, 4) comes from adding the scale factor to each coordinate instead of multiplying; (6, 1) comes from multiplying only the x-coordinate by 3 and leaving the y-coordinate unchanged; (2, 3) comes from multiplying only the y-coordinate by 3 and leaving the x-coordinate unchanged.)
- (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) 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.
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