High School Maths · Year 2 · Unit 12 · Lesson 4 of 5

Rotations and Enlargements: Turn It, Grow It, Shrink It, Flip It

About 15 minutes · rotating about the origin and about any centre (angle and direction) · finding the centre of a rotation · enlarging with positive, fractional and negative scale factors · finding the centre and scale factor from coordinates · areas after an enlargement · no calculator · exit quiz

Hook

In a pinhole camera, light from the tip of a candle at (3, 2) passes through a tiny hole at the origin and lands at (−6, −4).

The picture is bigger and upside down. What single transformation is this?

Each point goes through the hole and comes out on the other side, twice as far away: (3, 2) → (−6, −4). That is an enlargement with scale factor −2, centre the origin. A negative scale factor puts the image on the other side of the centre, upside down. In Unit 9 you met scale factors for similar shapes; here you will use them on a grid.

Learn

Master the essentials

A quarter turn clockwise about O: (x, y) goes to (y, −x).

Rotation

To describe a rotation, give the centre, the angle and the direction (clockwise or anticlockwise).

90° clockwise about O

90° clockwise about O

(x, y) goes to (y, −x).

(3, 5) → (5, −3)

90° anticlockwise about O

(x, y) goes to (−y, x).

(4, −1) → (1, 4)

180° about O

(x, y) goes to (−x, −y). The direction does not matter.

(−3, 6) → (3, −6)

Any centre

Find the movement from the centre to the point, turn that movement, then add it back on to the centre.

90° anticlockwise about (1, 2): (4, 2) → (1, 5)

Enlargement

Scale factor k, centre C: multiply the vector from C to each point by k. Lengths are multiplied by k and areas by k².

Centre (1, 1), k = 3: (3, 2) → (7, 4)

Fractional k

If k is between 0 and 1, the image is smaller and nearer to the centre.

k = ½, centre O: (6, −4) → (3, −2)

Negative k

The image is on the other side of the centre and upside down. Its size uses the number without the sign.

k = −2, centre O: (3, −1) → (−6, 2)

Worked examples

See it done

1. Triangle R has corners (1, 1), (3, 1) and (3, 2). Rotate R through 90° anticlockwise about the point (−1, 0). Find the corners of the image.

  1. From the centre (−1, 0) to (1, 1) is 21.
  2. A quarter turn anticlockwise sends 21 to −12. So (1, 1) → (−1 − 1, 0 + 2) = (−2, 2).
  3. In the same way, (3, 1): 41 → −14, giving (−2, 4); (3, 2): 42 → −24, giving (−3, 4).

(−2, 2), (−2, 4) and (−3, 4)

2. Triangle E has corners (4, 3), (6, 3) and (4, 7). Enlarge E by scale factor −½, centre (2, 1). Find the corners of the image.

  1. From the centre (2, 1) to (4, 3) is 22. Multiply by −½: −1−1, so (4, 3) → (1, 0).
  2. (6, 3): 42 → −2−1, giving (0, 0).
  3. (4, 7): 26 → −1−3, giving (1, −2).
  4. The image is half the size, on the other side of the centre and upside down.

(1, 0), (0, 0) and (1, −2)

3. Triangle G has corners (1, 1), (2, 1) and (1, 2). Triangle H has corners (5, 3), (8, 3) and (5, 6). Describe fully the single transformation that maps G onto H.

  1. H is the same way up and bigger: an enlargement with a positive scale factor.
  2. The side from (1, 1) to (2, 1) has length 1; on H it is from (5, 3) to (8, 3), length 3. So k = 3.
  3. The centre C satisfies (5, 3) = C + 3((1, 1) − C), so 5 = 3 − 2x and 3 = 3 − 2y. That gives C = (−1, 0).
  4. Check: (−1, 0) + 3 × 22 = (5, 6) ✓.

Enlargement, scale factor 3, centre (−1, 0)

Common mistake: Turning the wrong way, or forgetting the centre. 90° clockwise about O sends (2, 5) to (5, −2), not (−5, 2). When the centre is not the origin, turn the movement from the centre, then add the centre back on. And a negative scale factor does not make a shape smaller: k = −3 makes it three times as big, on the other side of the centre.

PractiseNo calculator

10 questions with instant feedback

Choose the correct answer. Always work from the centre: find the vector from the centre to the point, turn it or multiply it, then add it back on. Questions 1–3 are warm-ups, 4–7 are at exam standard and 8–10 are challenge questions. After you answer, the feedback shows the full solution.

Score: 0 / 10Answered: 0 / 10

Warm-up (The basic skill: one or two steps.)

1.Rotate the point (6, 2) through 90° clockwise about the origin.

Not quite. This is a 90° anticlockwise turn.

Solution

  1. 90° clockwise about O: (x, y) → (y, −x).
  2. (6, 2) → (2, −6).

Answer: D. (2, −6).

Not quite. This is a half turn (180°).

Solution

  1. 90° clockwise about O: (x, y) → (y, −x).
  2. (6, 2) → (2, −6).

Answer: D. (2, −6).

Not quite. This only swaps the coordinates. That is a reflection in y = x.

Solution

  1. 90° clockwise about O: (x, y) → (y, −x).
  2. (6, 2) → (2, −6).

Answer: D. (2, −6).

Correct.

Solution

  1. 90° clockwise about O: (x, y) → (y, −x).
  2. (6, 2) → (2, −6).

2.Enlarge the point (−3, 4) by scale factor 3, centre the origin.

Not quite. This adds 3. An enlargement multiplies.

Solution

  1. With centre O, multiply both coordinates by 3.
  2. (−3 × 3, 4 × 3) = (−9, 12).

Answer: C. (−9, 12).

Not quite. This uses scale factor −3. The scale factor is +3.

Solution

  1. With centre O, multiply both coordinates by 3.
  2. (−3 × 3, 4 × 3) = (−9, 12).

Answer: C. (−9, 12).

Correct.

Solution

  1. With centre O, multiply both coordinates by 3.
  2. (−3 × 3, 4 × 3) = (−9, 12).

Not quite. This multiplies only the x-coordinate.

Solution

  1. With centre O, multiply both coordinates by 3.
  2. (−3 × 3, 4 × 3) = (−9, 12).

Answer: C. (−9, 12).

3.An enlargement with centre the origin maps (10, −6) onto (5, −3). Find the scale factor.

Not quite. This divides the object by the image. Scale factor = image ÷ object.

Solution

  1. Scale factor = image ÷ object.
  2. 5 ÷ 10 = ½ (and −3 ÷ −6 = ½ too).

Answer: C. 0.5.

Not quite. This is the difference 10 − 5. A scale factor is a multiplier.

Solution

  1. Scale factor = image ÷ object.
  2. 5 ÷ 10 = ½ (and −3 ÷ −6 = ½ too).

Answer: C. 0.5.

Correct.

Solution

  1. Scale factor = image ÷ object.
  2. 5 ÷ 10 = ½ (and −3 ÷ −6 = ½ too).

Not quite. The image is on the same side of O as the object, so the scale factor is positive.

Solution

  1. Scale factor = image ÷ object.
  2. 5 ÷ 10 = ½ (and −3 ÷ −6 = ½ too).

Answer: C. 0.5.

Exam standard (Several steps, like real exam questions.)

4.Rotate the point (5, 4) through 90° anticlockwise about the point (2, 1).

Not quite. This turns about the origin. The centre is (2, 1).

Solution

  1. From the centre (2, 1) to (5, 4) is 33.
  2. A quarter turn anticlockwise: 33 → −33.
  3. Add to the centre: (2 − 3, 1 + 3) = (−1, 4).

Answer: B. (−1, 4).

Correct.

Solution

  1. From the centre (2, 1) to (5, 4) is 33.
  2. A quarter turn anticlockwise: 33 → −33.
  3. Add to the centre: (2 − 3, 1 + 3) = (−1, 4).

Not quite. This turns clockwise.

Solution

  1. From the centre (2, 1) to (5, 4) is 33.
  2. A quarter turn anticlockwise: 33 → −33.
  3. Add to the centre: (2 − 3, 1 + 3) = (−1, 4).

Answer: B. (−1, 4).

Not quite. This is the turned movement. Add it back on to the centre.

Solution

  1. From the centre (2, 1) to (5, 4) is 33.
  2. A quarter turn anticlockwise: 33 → −33.
  3. Add to the centre: (2 − 3, 1 + 3) = (−1, 4).

Answer: B. (−1, 4).

5.Enlarge the point (9, −3) by scale factor 13, centre (3, 0).

Correct.

Solution

  1. From (3, 0) to (9, −3) is 6−3.
  2. Multiply by 13: 2−1.
  3. Add to the centre: (3 + 2, 0 − 1) = (5, −1).

Not quite. This uses the origin as the centre. The centre is (3, 0).

Solution

  1. From (3, 0) to (9, −3) is 6−3.
  2. Multiply by 13: 2−1.
  3. Add to the centre: (3 + 2, 0 − 1) = (5, −1).

Answer: A. (5, −1).

Not quite. This is the new movement from the centre. Add it back on to (3, 0).

Solution

  1. From (3, 0) to (9, −3) is 6−3.
  2. Multiply by 13: 2−1.
  3. Add to the centre: (3 + 2, 0 − 1) = (5, −1).

Answer: A. (5, −1).

Not quite. This multiplies by 3. The scale factor is one third.

Solution

  1. From (3, 0) to (9, −3) is 6−3.
  2. Multiply by 13: 2−1.
  3. Add to the centre: (3 + 2, 0 − 1) = (5, −1).

Answer: A. (5, −1).

6.Point A is at (4, 1). Find its image after an enlargement with scale factor −2, centre (1, −1).

Not quite. This uses +2. A negative scale factor puts the image on the other side of the centre.

Solution

  1. From (1, −1) to (4, 1) is 32.
  2. Multiply by −2: −6−4.
  3. Add to the centre: (1 − 6, −1 − 4) = (−5, −5).

Answer: D. (−5, −5).

Not quite. This uses the origin as the centre.

Solution

  1. From (1, −1) to (4, 1) is 32.
  2. Multiply by −2: −6−4.
  3. Add to the centre: (1 − 6, −1 − 4) = (−5, −5).

Answer: D. (−5, −5).

Not quite. This is the new movement from the centre. Add it back on to (1, −1).

Solution

  1. From (1, −1) to (4, 1) is 32.
  2. Multiply by −2: −6−4.
  3. Add to the centre: (1 − 6, −1 − 4) = (−5, −5).

Answer: D. (−5, −5).

Correct.

Solution

  1. From (1, −1) to (4, 1) is 32.
  2. Multiply by −2: −6−4.
  3. Add to the centre: (1 − 6, −1 − 4) = (−5, −5).

7.The table shows three corners of a triangle and their images after an enlargement. Find the scale factor and the centre.

CornerImage
(3, 4)(7, 10)
(5, 4)(11, 10)
(3, 6)(7, 14)

Correct.

Solution

  1. Side from (3, 4) to (5, 4) has length 2; its image from (7, 10) to (11, 10) has length 4. So k = 2.
  2. The centre C: (7, 10) = C + 2((3, 4) − C) = 2(3, 4) − C, so C = (6 − 7, 8 − 10) = (−1, −2).
  3. Check (3, 6): (−1, −2) + 2 × 48 = (7, 14) ✓.

Not quite. This maps the image back onto the object. Scale factor = image length ÷ object length.

Solution

  1. Side from (3, 4) to (5, 4) has length 2; its image from (7, 10) to (11, 10) has length 4. So k = 2.
  2. The centre C: (7, 10) = C + 2((3, 4) − C) = 2(3, 4) − C, so C = (6 − 7, 8 − 10) = (−1, −2).
  3. Check (3, 6): (−1, −2) + 2 × 48 = (7, 14) ✓.

Answer: A. Enlargement, scale factor 2, centre (−1, −2).

Not quite. Sign slip: the centre C solves (7, 10) = 2(3, 4) − C, so C = (6 − 7, 8 − 10).

Solution

  1. Side from (3, 4) to (5, 4) has length 2; its image from (7, 10) to (11, 10) has length 4. So k = 2.
  2. The centre C: (7, 10) = C + 2((3, 4) − C) = 2(3, 4) − C, so C = (6 − 7, 8 − 10) = (−1, −2).
  3. Check (3, 6): (−1, −2) + 2 × 48 = (7, 14) ✓.

Answer: A. Enlargement, scale factor 2, centre (−1, −2).

Not quite. With centre O, (3, 4) would go to (6, 8), not (7, 10).

Solution

  1. Side from (3, 4) to (5, 4) has length 2; its image from (7, 10) to (11, 10) has length 4. So k = 2.
  2. The centre C: (7, 10) = C + 2((3, 4) − C) = 2(3, 4) − C, so C = (6 − 7, 8 − 10) = (−1, −2).
  3. Check (3, 6): (−1, −2) + 2 × 48 = (7, 14) ✓.

Answer: A. Enlargement, scale factor 2, centre (−1, −2).

Challenge (Harder problems that mix ideas or need reasoning.)

8.A rotation of 90° clockwise maps the point (1, 3) onto (5, 1). Find the centre of the rotation.

Correct.

Solution

  1. Let the centre be (c, d). The movement from the centre to (1, 3) is 1 − c3 − d.
  2. A clockwise quarter turn sends it to 3 − dc − 1, which must equal (5 − c, 1 − d).
  3. 3 − d = 5 − c and c − 1 = 1 − d. Adding: 2 = 6 − 2c, so c = 2 and d = 0.
  4. Check: from (2, 0) to (1, 3) is −13 → 31, giving (5, 1) ✓.

Not quite. This is the midpoint of the two points. That is the centre only for a half turn.

Solution

  1. Let the centre be (c, d). The movement from the centre to (1, 3) is 1 − c3 − d.
  2. A clockwise quarter turn sends it to 3 − dc − 1, which must equal (5 − c, 1 − d).
  3. 3 − d = 5 − c and c − 1 = 1 − d. Adding: 2 = 6 − 2c, so c = 2 and d = 0.
  4. Check: from (2, 0) to (1, 3) is −13 → 31, giving (5, 1) ✓.

Answer: A. (2, 0).

Not quite. About the origin, (1, 3) would go to (3, −1).

Solution

  1. Let the centre be (c, d). The movement from the centre to (1, 3) is 1 − c3 − d.
  2. A clockwise quarter turn sends it to 3 − dc − 1, which must equal (5 − c, 1 − d).
  3. 3 − d = 5 − c and c − 1 = 1 − d. Adding: 2 = 6 − 2c, so c = 2 and d = 0.
  4. Check: from (2, 0) to (1, 3) is −13 → 31, giving (5, 1) ✓.

Answer: A. (2, 0).

Not quite. This is the centre for a 90° anticlockwise turn.

Solution

  1. Let the centre be (c, d). The movement from the centre to (1, 3) is 1 − c3 − d.
  2. A clockwise quarter turn sends it to 3 − dc − 1, which must equal (5 − c, 1 − d).
  3. 3 − d = 5 − c and c − 1 = 1 − d. Adding: 2 = 6 − 2c, so c = 2 and d = 0.
  4. Check: from (2, 0) to (1, 3) is −13 → 31, giving (5, 1) ✓.

Answer: A. (2, 0).

9.Which working shows that an enlargement with scale factor −1, centre the origin, has the same effect as a rotation of 180° about the origin?

Not quite. Both coordinates are multiplied by −1. (−x, y) is a reflection in the y-axis.

Solution

  1. Enlargement, k = −1, centre O: multiply both coordinates by −1.
  2. (x, y) → (−x, −y).
  3. A half turn about O has the same rule ✓.

Answer: D. The enlargement sends (x, y) to (−1 × x, −1 × y) = (−x, −y), which is exactly the rule for a half turn about O.

Not quite. This adds −1. A scale factor multiplies.

Solution

  1. Enlargement, k = −1, centre O: multiply both coordinates by −1.
  2. (x, y) → (−x, −y).
  3. A half turn about O has the same rule ✓.

Answer: D. The enlargement sends (x, y) to (−1 × x, −1 × y) = (−x, −y), which is exactly the rule for a half turn about O.

Not quite. Neither changes the size: the length is multiplied by 1 (the size of −1).

Solution

  1. Enlargement, k = −1, centre O: multiply both coordinates by −1.
  2. (x, y) → (−x, −y).
  3. A half turn about O has the same rule ✓.

Answer: D. The enlargement sends (x, y) to (−1 × x, −1 × y) = (−x, −y), which is exactly the rule for a half turn about O.

Correct.

Solution

  1. Enlargement, k = −1, centre O: multiply both coordinates by −1.
  2. (x, y) → (−x, −y).
  3. A half turn about O has the same rule ✓.

10.An enlargement with scale factor −12, centre the origin, maps the point A onto (3, −4). Find the coordinates of A.

Not quite. This applies the enlargement again. To go back, divide by −½: multiply by −2.

Solution

  1. Image = −½ × A, so A = image ÷ (−½) = −2 × image.
  2. A = (−2 × 3, −2 × (−4)) = (−6, 8).
  3. Check: −½ × (−6, 8) = (3, −4) ✓.

Answer: B. (−6, 8).

Correct.

Solution

  1. Image = −½ × A, so A = image ÷ (−½) = −2 × image.
  2. A = (−2 × 3, −2 × (−4)) = (−6, 8).
  3. Check: −½ × (−6, 8) = (3, −4) ✓.

Not quite. This multiplies by 2. Dividing by −½ means multiplying by −2.

Solution

  1. Image = −½ × A, so A = image ÷ (−½) = −2 × image.
  2. A = (−2 × 3, −2 × (−4)) = (−6, 8).
  3. Check: −½ × (−6, 8) = (3, −4) ✓.

Answer: B. (−6, 8).

Not quite. This only changes the signs. A is twice as far from O as its image.

Solution

  1. Image = −½ × A, so A = image ÷ (−½) = −2 × image.
  2. A = (−2 × 3, −2 × (−4)) = (−6, 8).
  3. Check: −½ × (−6, 8) = (3, −4) ✓.

Answer: B. (−6, 8).

Apply

Turn it, then flip it

Triangle A has corners (1, 1), (3, 1) and (1, 2).

1. Rotate A through 90° clockwise about (−1, −1). Write down the corners of the image B.

2. Enlarge A by scale factor −2, centre the origin. Write down the corners of the image C.

3. How many times bigger is the area of C than the area of A?

  1. From (−1, −1) to (1, 1) is 22; clockwise this becomes 2−2, so (1, 1) → (1, −3).
  2. (3, 1): 42 → 2−4, giving (1, −5). (1, 2): 23 → 3−2, giving (2, −3).
  3. B: (1, −3), (1, −5) and (2, −3).
  1. Multiply every coordinate by −2.
  2. C: (−2, −2), (−6, −2) and (−2, −4).
  1. Lengths are multiplied by 2 (the size of −2), so areas are multiplied by 2² = 4.
  2. Check: A has area ½ × 2 × 1 = 1 and C has area ½ × 4 × 2 = 4. The area is 4 times as big.

Next: the Lesson 4 exit quiz. Five questions, recorded on your dashboard. Aim for 4 out of 5.

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