High School Maths · Year 2 · Unit 12 · Lesson 3 of 5

Translations and Reflections: Slide It, Flip It, Describe It Fully

About 15 minutes · translating by a column vector · finding the vector of a translation · reflecting in vertical and horizontal lines such as x = 2 and y = −1 · reflecting in y = x and y = −x · finding a mirror line from a point and its image · describing a single transformation fully · no calculator · exit quiz

Hook

A footprint is at the point P(6, 1). Its reflection in a mirror is at P′(−2, 1).

Where is the mirror line?

The mirror is exactly halfway between a point and its image. Halfway between x = 6 and x = −2 is (6 + (−2)) ÷ 2 = 2, so the mirror is the line x = 2. Finding the mirror from a point and its image is a key exam skill in this lesson.

Learn

Master the essentials

Reflecting in y = x swaps the coordinates: (4, −1) goes to (−1, 4).

Object and image

The object is the shape you start with; the image is where it ends up. A point A goes to A′. These moves keep the size and shape.

Object A, image A′ (congruent)

Translation

Translate by a column vector: add the top number to x and the bottom number to y. Every point moves the same way.

(2, −3) by −54 goes to (−3, 1)

Finding the vector

For a translation, the vector is image minus object, for any one point.

(1, 6) → (4, 2): 3−4

Mirror x = k

A vertical mirror. y stays the same; the point ends up the same distance on the other side.

In x = 2, (5, 1) → (−1, 1)

Mirror y = k

A horizontal mirror. x stays the same; count squares to the mirror, then the same again.

In y = −1, (3, 4) → (3, −6)

Mirror y = x

Swap the two coordinates.

(5, −2) → (−2, 5)

Mirror y = −x

Swap the coordinates and change both signs.

(6, 1) → (−1, −6)

Describe it fully

Translation: give the column vector. Reflection: give the equation of the mirror line, which passes through the midpoint of every point and its image.

Reflection in the line y = 3

Worked examples

See it done

1. Triangle T has corners (1, 2), (4, 2) and (1, 4). Translate T by the vector −6−3. Find the corners of the image T′.

  1. Add −6 to every x-coordinate and −3 to every y-coordinate.
  2. (1, 2) → (1 − 6, 2 − 3) = (−5, −1).
  3. (4, 2) → (−2, −1) and (1, 4) → (−5, 1).

(−5, −1), (−2, −1) and (−5, 1)

2. Triangle S has corners (1, 2), (3, 2) and (3, 5). Reflect S in the line y = −x. Find the corners of the image S′.

  1. In y = −x, swap the coordinates and change both signs: (x, y) → (−y, −x).
  2. (1, 2) → (−2, −1).
  3. (3, 2) → (−2, −3) and (3, 5) → (−5, −3).

(−2, −1), (−2, −3) and (−5, −3)

3. Triangle A has corners (2, 1), (4, 1) and (4, 2). Triangle B has corners (2, −5), (4, −5) and (4, −6). Describe fully the single transformation that maps A onto B.

  1. B is flipped over (upside down), so this is a reflection, not a translation.
  2. Each x stays the same, so the mirror is a horizontal line y = k.
  3. Midpoint of (2, 1) and (2, −5): y = (1 + (−5)) ÷ 2 = −2. Check: (4, 2) and (4, −6) give (2 + (−6)) ÷ 2 = −2 ✓.

Reflection in the line y = −2

Common mistake: Reflecting in y = −1 by just changing the sign of y. That only works for the x-axis (y = 0). Count squares instead: (3, 4) is 5 squares above y = −1, so its image is 5 squares below, at (3, −6). Also, "describe fully" means giving the vector or the mirror line, not just the word.

PractiseNo calculator

10 questions with instant feedback

Choose the correct answer. For a reflection, find the mirror line first; for a translation, find the vector. Check your answer with a second point. 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.The point (−4, 3) is translated by the vector 6−5. Find the coordinates of its image.

Correct.

Solution

  1. Add the top number to x: −4 + 6 = 2.
  2. Add the bottom number to y: 3 + (−5) = −2.
  3. The image is (2, −2).

Not quite. This subtracts the vector. A translation adds it.

Solution

  1. Add the top number to x: −4 + 6 = 2.
  2. Add the bottom number to y: 3 + (−5) = −2.
  3. The image is (2, −2).

Answer: A. (2, −2).

Not quite. This adds 5 to the y-coordinate. The bottom number is −5.

Solution

  1. Add the top number to x: −4 + 6 = 2.
  2. Add the bottom number to y: 3 + (−5) = −2.
  3. The image is (2, −2).

Answer: A. (2, −2).

Not quite. This adds the top number to y and the bottom number to x. The top number goes with x.

Solution

  1. Add the top number to x: −4 + 6 = 2.
  2. Add the bottom number to y: 3 + (−5) = −2.
  3. The image is (2, −2).

Answer: A. (2, −2).

2.Reflect the point (5, −3) in the line y = x.

Not quite. This is the reflection in y = −x. In y = x, just swap.

Solution

  1. In y = x, swap the coordinates.
  2. (5, −3) → (−3, 5).

Answer: B. (−3, 5).

Correct.

Solution

  1. In y = x, swap the coordinates.
  2. (5, −3) → (−3, 5).

Not quite. This is the reflection in the y-axis.

Solution

  1. In y = x, swap the coordinates.
  2. (5, −3) → (−3, 5).

Answer: B. (−3, 5).

Not quite. This is the reflection in the x-axis.

Solution

  1. In y = x, swap the coordinates.
  2. (5, −3) → (−3, 5).

Answer: B. (−3, 5).

3.Shape P is translated onto shape Q. The corner (−2, 4) of P moves to (3, −1) on Q. Which column vector describes the translation?

Not quite. This is the translation from Q back to P. Use image minus object.

Solution

  1. Vector = image − object.
  2. Top: 3 − (−2) = 5.
  3. Bottom: −1 − 4 = −5.

Answer: C. 5−5.

Not quite. This adds the coordinates. Subtract: image minus object.

Solution

  1. Vector = image − object.
  2. Top: 3 − (−2) = 5.
  3. Bottom: −1 − 4 = −5.

Answer: C. 5−5.

Correct.

Solution

  1. Vector = image − object.
  2. Top: 3 − (−2) = 5.
  3. Bottom: −1 − 4 = −5.

Not quite. The bottom number is −1 − 4 = −5. Watch the sign.

Solution

  1. Vector = image − object.
  2. Top: 3 − (−2) = 5.
  3. Bottom: −1 − 4 = −5.

Answer: C. 5−5.

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

4.Triangle T has corners (1, 1), (4, 1) and (4, 3). It is reflected in the line x = −1. Which list gives the corners of the image?

Not quite. This reflects in the y-axis (x = 0), not in x = −1.

Solution

  1. The mirror x = −1 is vertical, so each y stays the same.
  2. (1, 1) is 2 squares right of x = −1, so its image is 2 squares left: (−3, 1).
  3. (4, 1) is 5 squares right, so (−6, 1); (4, 3) goes to (−6, 3).

Answer: C. (−3, 1), (−6, 1) and (−6, 3).

Not quite. This reflects in y = −1, a horizontal line. x = −1 is vertical.

Solution

  1. The mirror x = −1 is vertical, so each y stays the same.
  2. (1, 1) is 2 squares right of x = −1, so its image is 2 squares left: (−3, 1).
  3. (4, 1) is 5 squares right, so (−6, 1); (4, 3) goes to (−6, 3).

Answer: C. (−3, 1), (−6, 1) and (−6, 3).

Correct.

Solution

  1. The mirror x = −1 is vertical, so each y stays the same.
  2. (1, 1) is 2 squares right of x = −1, so its image is 2 squares left: (−3, 1).
  3. (4, 1) is 5 squares right, so (−6, 1); (4, 3) goes to (−6, 3).

Not quite. This changes the sign of x and takes 1 away. Each point must end up as far beyond x = −1 as it started.

Solution

  1. The mirror x = −1 is vertical, so each y stays the same.
  2. (1, 1) is 2 squares right of x = −1, so its image is 2 squares left: (−3, 1).
  3. (4, 1) is 5 squares right, so (−6, 1); (4, 3) goes to (−6, 3).

Answer: C. (−3, 1), (−6, 1) and (−6, 3).

5.Triangle A has corners (1, 3), (4, 3) and (4, 5). Triangle B has corners (−3, −1), (−3, −4) and (−5, −4). Describe fully the single transformation that maps A onto B.

Not quite. Reflecting in y = x would send (1, 3) to (3, 1).

Solution

  1. Test (x, y) → (−y, −x): (1, 3) → (−3, −1) ✓.
  2. (4, 3) → (−3, −4) ✓ and (4, 5) → (−5, −4) ✓.
  3. Swap and change both signs: a reflection in y = −x.

Answer: D. Reflection in the line y = −x.

Not quite. A half turn would send (1, 3) to (−1, −3).

Solution

  1. Test (x, y) → (−y, −x): (1, 3) → (−3, −1) ✓.
  2. (4, 3) → (−3, −4) ✓ and (4, 5) → (−5, −4) ✓.
  3. Swap and change both signs: a reflection in y = −x.

Answer: D. Reflection in the line y = −x.

Not quite. This moves (1, 3) correctly, but (4, 3) would go to (0, −1), not (−3, −4).

Solution

  1. Test (x, y) → (−y, −x): (1, 3) → (−3, −1) ✓.
  2. (4, 3) → (−3, −4) ✓ and (4, 5) → (−5, −4) ✓.
  3. Swap and change both signs: a reflection in y = −x.

Answer: D. Reflection in the line y = −x.

Correct.

Solution

  1. Test (x, y) → (−y, −x): (1, 3) → (−3, −1) ✓.
  2. (4, 3) → (−3, −4) ✓ and (4, 5) → (−5, −4) ✓.
  3. Swap and change both signs: a reflection in y = −x.

6.The table shows where three corners of shape S go under a single transformation. Describe the transformation fully.

Point on SImage on S′
(2, 5)(2, −3)
(4, 5)(4, −3)
(4, 7)(4, −5)

Not quite. The x-axis would send (2, 5) to (2, −5).

Solution

  1. Each x stays the same, so the mirror is a horizontal line.
  2. Midpoint of (2, 5) and (2, −3): y = (5 + (−3)) ÷ 2 = 1.
  3. Check (4, 7) and (4, −5): (7 + (−5)) ÷ 2 = 1 ✓.

Answer: D. Reflection in the line y = 1.

Not quite. This fits the first two points, but (4, 7) would go to (4, −1), not (4, −5).

Solution

  1. Each x stays the same, so the mirror is a horizontal line.
  2. Midpoint of (2, 5) and (2, −3): y = (5 + (−3)) ÷ 2 = 1.
  3. Check (4, 7) and (4, −5): (7 + (−5)) ÷ 2 = 1 ✓.

Answer: D. Reflection in the line y = 1.

Not quite. This halves the gap 5 − (−3) = 8. The mirror is at the midpoint: (5 + (−3)) ÷ 2.

Solution

  1. Each x stays the same, so the mirror is a horizontal line.
  2. Midpoint of (2, 5) and (2, −3): y = (5 + (−3)) ÷ 2 = 1.
  3. Check (4, 7) and (4, −5): (7 + (−5)) ÷ 2 = 1 ✓.

Answer: D. Reflection in the line y = 1.

Correct.

Solution

  1. Each x stays the same, so the mirror is a horizontal line.
  2. Midpoint of (2, 5) and (2, −3): y = (5 + (−3)) ÷ 2 = 1.
  3. Check (4, 7) and (4, −5): (7 + (−5)) ÷ 2 = 1 ✓.

7.Point P is reflected in the line x = 4. Its image is (9, −2). Find the coordinates of P.

Not quite. This reflects in the y-axis. The mirror is x = 4.

Solution

  1. A reflection undoes itself, so reflect (9, −2) in x = 4.
  2. 9 is 5 squares to the right of x = 4.
  3. P is 5 squares to the left: x = 4 − 5 = −1, so P = (−1, −2).

Answer: C. (−1, −2).

Not quite. This moves 4 squares. (9, −2) is 5 squares from x = 4, so P is 5 squares the other side.

Solution

  1. A reflection undoes itself, so reflect (9, −2) in x = 4.
  2. 9 is 5 squares to the right of x = 4.
  3. P is 5 squares to the left: x = 4 − 5 = −1, so P = (−1, −2).

Answer: C. (−1, −2).

Correct.

Solution

  1. A reflection undoes itself, so reflect (9, −2) in x = 4.
  2. 9 is 5 squares to the right of x = 4.
  3. P is 5 squares to the left: x = 4 − 5 = −1, so P = (−1, −2).

Not quite. This moves away from the mirror. P is on the other side of x = 4 from its image.

Solution

  1. A reflection undoes itself, so reflect (9, −2) in x = 4.
  2. 9 is 5 squares to the right of x = 4.
  3. P is 5 squares to the left: x = 4 − 5 = −1, so P = (−1, −2).

Answer: C. (−1, −2).

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

8.The point A(1, 8) is reflected onto A′(7, 2). Find the equation of the mirror line.

Correct.

Solution

  1. The mirror passes through the midpoint of AA′: ((1 + 7) ÷ 2, (8 + 2) ÷ 2) = (4, 5).
  2. AA′ has gradient (2 − 8) ÷ (7 − 1) = −1, so the mirror has gradient 1.
  3. y = x + c through (4, 5): 5 = 4 + c, so c = 1. The mirror is y = x + 1.

Not quite. The mirror must pass through the midpoint (4, 5). y = x does not.

Solution

  1. The mirror passes through the midpoint of AA′: ((1 + 7) ÷ 2, (8 + 2) ÷ 2) = (4, 5).
  2. AA′ has gradient (2 − 8) ÷ (7 − 1) = −1, so the mirror has gradient 1.
  3. y = x + c through (4, 5): 5 = 4 + c, so c = 1. The mirror is y = x + 1.

Answer: A. y = x + 1.

Not quite. This is the line through A and A′. The mirror is perpendicular to it.

Solution

  1. The mirror passes through the midpoint of AA′: ((1 + 7) ÷ 2, (8 + 2) ÷ 2) = (4, 5).
  2. AA′ has gradient (2 − 8) ÷ (7 − 1) = −1, so the mirror has gradient 1.
  3. y = x + c through (4, 5): 5 = 4 + c, so c = 1. The mirror is y = x + 1.

Answer: A. y = x + 1.

Not quite. A vertical mirror keeps y the same, but y changes from 8 to 2.

Solution

  1. The mirror passes through the midpoint of AA′: ((1 + 7) ÷ 2, (8 + 2) ÷ 2) = (4, 5).
  2. AA′ has gradient (2 − 8) ÷ (7 − 1) = −1, so the mirror has gradient 1.
  3. y = x + c through (4, 5): 5 = 4 + c, so c = 1. The mirror is y = x + 1.

Answer: A. y = x + 1.

9.Which working shows that the line x = 3 is the mirror line that maps (−1, 4) onto (7, 4)?

Not quite. This halves the distance between the points, 8 ÷ 2 = 4, but the mirror is at the midpoint: (−1 + 7) ÷ 2 = 3.

Solution

  1. The mirror goes through the midpoint of a point and its image: (3, 4).
  2. (3, 4) is on the line x = 3.
  3. The join is horizontal, at right angles to x = 3 ✓.

Answer: B. The midpoint is ((−1 + 7) ÷ 2, 4) = (3, 4), which lies on x = 3, and the line joining the points is horizontal, so it is perpendicular to x = 3.

Correct.

Solution

  1. The mirror goes through the midpoint of a point and its image: (3, 4).
  2. (3, 4) is on the line x = 3.
  3. The join is horizontal, at right angles to x = 3 ✓.

Not quite. Points with the same y are reflected in a vertical line x = k, not in y = 4.

Solution

  1. The mirror goes through the midpoint of a point and its image: (3, 4).
  2. (3, 4) is on the line x = 3.
  3. The join is horizontal, at right angles to x = 3 ✓.

Answer: B. The midpoint is ((−1 + 7) ÷ 2, 4) = (3, 4), which lies on x = 3, and the line joining the points is horizontal, so it is perpendicular to x = 3.

Not quite. The midpoint has x = 3, so the mirror is the vertical line x = 3, not y = 3.

Solution

  1. The mirror goes through the midpoint of a point and its image: (3, 4).
  2. (3, 4) is on the line x = 3.
  3. The join is horizontal, at right angles to x = 3 ✓.

Answer: B. The midpoint is ((−1 + 7) ÷ 2, 4) = (3, 4), which lies on x = 3, and the line joining the points is horizontal, so it is perpendicular to x = 3.

10.A translation maps (1, 4) onto (−2, 9). The same translation maps (2p, 5) onto (p + 3, q). Find p and q.

Not quite. p + 3 − 2p = −3 gives −p = −6. The 3 must be taken away from both sides, not added.

Solution

  1. The vector is image − object: −35.
  2. Tops: (p + 3) − 2p = −3, so −p = −6 and p = 6.
  3. Bottoms: q − 5 = 5, so q = 10.

Answer: D. p = 6, q = 10.

Not quite. From −p = −6, divide by −1: p = 6.

Solution

  1. The vector is image − object: −35.
  2. Tops: (p + 3) − 2p = −3, so −p = −6 and p = 6.
  3. Bottoms: q − 5 = 5, so q = 10.

Answer: D. p = 6, q = 10.

Not quite. This uses the vector the wrong way round: q − 5 = +5, not −5.

Solution

  1. The vector is image − object: −35.
  2. Tops: (p + 3) − 2p = −3, so −p = −6 and p = 6.
  3. Bottoms: q − 5 = 5, so q = 10.

Answer: D. p = 6, q = 10.

Correct.

Solution

  1. The vector is image − object: −35.
  2. Tops: (p + 3) − 2p = −3, so −p = −6 and p = 6.
  3. Bottoms: q − 5 = 5, so q = 10.

Apply

Design a pattern

A designer draws triangle A with corners (1, 1), (3, 1) and (1, 4), and copies it with transformations.

1. Translate A by the vector −52. Write down the corners of the image B.

2. Reflect A in the line y = −x. Write down the corners of the image C.

3. Describe fully the single transformation that maps C back onto A.

  1. Add −5 to each x and 2 to each y.
  2. B: (−4, 3), (−2, 3) and (−4, 6).
  1. In y = −x, (x, y) → (−y, −x).
  2. C: (−1, −1), (−1, −3) and (−4, −1).
  1. Reflecting C in y = −x again: (−1, −1) → (1, 1), (−1, −3) → (3, 1), (−4, −1) → (1, 4).
  2. That is triangle A, so the transformation is a reflection in the line y = −x. A reflection undoes itself.

Next: the Lesson 3 exit quiz. Five questions, recorded on your dashboard. Aim for 4 out of 5.

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