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.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.
- 2.Point B is at (3, 5). It is reflected in the line y = 2. Work out the coordinates of the image of point B.
- 3.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.
- 4.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.
- 5.Shape S has an area of 5 cm². Shape S is enlarged by a scale factor of −3 to give shape T. Work out the area of shape T.
- 6.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.
- 7.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.
- 8.Point A(1, 2) is enlarged to give image point A′(7, −4). The centre of enlargement lies on the x-axis. Work out the scale factor of the enlargement.
- 9.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?
- 10.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
- 11.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.
- 12.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.
- 13.A photograph is 10 cm wide and 15 cm tall. It is enlarged to make a similar poster that is 40 cm wide. The poster costs £0.80 per centimetre of its height to print, based on its full height. Work out the cost of printing the poster.
- 14.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.
- 15.An enlargement has a scale factor of −2. Priti says the image must be a mirror image of the original shape. Is Priti correct?
Answer key
- (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).
- (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.)
- (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.
- (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.
- (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.
- (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.
- (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) (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).
- (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.
- (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.)
- (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) 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.
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