GCSE Maths revision · Higher: geometry

Combined transformations

Higher only

The idea

Do the first transformation, then the second on the result. The order matters. An invariant point stays in the same place after the transformation.

  1. Draw the image after the first transformation.
  2. Transform that image with the second one.
  3. Describe the single transformation from the original to the final shape, if asked.

Watch out Reflect then rotate does not usually give the same result as rotate then reflect.

See it worked

Worked exampleNon-calculator3 marks

Shape SS is reflected in the line y=xy = x to give shape S1S_1. S1S_1 is then reflected in the xx-axis to give shape S2S_2.

Describe fully the single transformation that maps SS onto S2S_2.

  1. Follow a general point: (x,y)→(y,x)(x, y) \to (y, x) after the first reflection.
  2. Then (y,x)→(y,−x)(y, x) \to (y, -x) after reflecting in the xx-axis.
  3. (x,y)→(y,−x)(x, y) \to (y, -x) is a rotation of 90∘90^\circ clockwise about the origin. Check: (1,0)→(0,−1)(1, 0) \to (0, -1).

Answer Rotation 90∘90^\circ clockwise about (0,0)(0, 0)

Your turn

Question 1Non-calculator3 marks

Point P(−2-2, 4) is reflected in the line x = 1. Its image is then reflected in the line x = 5. Work out the coordinates of the final image.

Write your answer, then show the working
  1.  2×1−(−2)2\times 1-(-2)
  2.  2×5−42\times 5-4
  3. Reflection in x = a sends x to 2a −- x and leaves y unchanged.
  4. The first image is (2 ×\times 1 −- (−2-2), 4) = (4, 4).
  5. The final image is (2 ×\times 5 −- 4, 4) = (6, 4).

Answer (6, 4)

Question 2Non-calculator5 marks

A point P(3,1)(3,1) is rotated 90 degrees clockwise about the origin and then translated by (3,−2)(3,-2).

(a) Find the final image of P.

(b) Find the final image if the order is reversed.

Question 3Non-calculator3 marks

A shape is reflected in the line x=ax = a and then its image is reflected in the line x=bx = b.

Prove that the combined transformation is a translation by the vector with components 2(b−a)2(b - a) and 0.

Write your answer, then show the working
  1. Reflecting in x=ax = a keeps yy and sends xx to the same distance on the other side: x→2a−xx \to 2a - x.
  2. Reflecting that in x=bx = b: 2a−x→2b−(2a−x)=x+2(b−a)2a - x \to 2b - (2a - x) = x + 2(b - a).
  3. So (x,y)→(x+2(b−a),  y)(x, y) \to (x + 2(b - a),\; y) for every point: a translation by 2(b−a)2(b - a) across and 0 up.

Answer (x,y)→(2a−x,y)→(2b−2a+x,y)(x, y) \to (2a - x, y) \to (2b - 2a + x, y)

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