Hard GCSE Transformations Questions

Challenging, exam-style GCSE Transformations questions with worked solutions. Stretch yourself on the hardest rotation, three quarter turn, centre a given point, anticlockwise problems.

rotationthree quarter turncentre a given pointanticlockwiseenlargementnegative scale factor
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
Triangle ABCABC is enlarged by scale factor −2-2, centre (1,1)(1, 1) to give triangle A′B′C′A'B'C'. The image of vertex BB is the point B′(−3,−1)B'(-3, -1). Find the coordinates of the original point BB.
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Worked solution

  1. Identify the transformation

    enlargement,k=−2,(1,1)\text{enlargement}, \quad k = -2, \quad (1, 1)

    The question describes a enlargement with scale factor -2, centre (1, 1), but this time the IMAGE is known and the original point is not.

  2. Write down the rule for the transformation

    (x, y)↦(1−2(x−1), 1−2(y−1))(x,\ y) \mapsto (1 - 2(x - 1),\ 1 - 2(y - 1))

    This rule sends the original point to the image. To go backwards it has to be undone.

  3. Say what undoing the transformation means

    B↦B′(−3,−1)⇒B′↦BB \mapsto B'(-3, -1) \Rightarrow B' \mapsto B

    Reversing a transformation means applying the transformation that takes the image back to the object.

  4. Write down the inverse transformation

    enlargement,k=−12,(1,1)\text{enlargement}, \quad k = -\frac{1}{2}, \quad (1, 1)

    The inverse of a enlargement with scale factor -2, centre (1, 1) is a enlargement with scale factor -1/2, centre (1, 1).

  5. Write down the rule for the inverse

    (x, y)↦(1−12(x−1), 1−12(y−1))(x,\ y) \mapsto (1 - \frac{1}{2}(x - 1),\ 1 - \frac{1}{2}(y - 1))

    This is the rule that will be applied to the image point.

  6. Substitute the image point into the inverse rule

    B′:(−4, −2)×(−12)=(2, 1)⇒B′′(3,2)B': (-4,\ -2) \times \left(-\frac{1}{2}\right) = (2,\ 1) \Rightarrow B''(3, 2)

    Putting B′(−3,−1)B'(-3, -1) into the inverse rule gives the original point.

  7. Write down the original point

    B=(3,2)B = (3, 2)

    So the vertex that was transformed is B(3,2)B(3, 2).

  8. Check by applying the original transformation

    B:(2, 1)×(−2)=(−4, −2)⇒B′(−3,−1)B: (2,\ 1) \times \left(-2\right) = (-4,\ -2) \Rightarrow B'(-3, -1)

    Transforming B(3,2)B(3, 2) forwards gives (−3,−1)(-3, -1), which is the image the question gave — the answer is right.

  9. Check the distance from the key line or point

    centre (1,1),k=−2\text{centre } (1, 1), \quad k = -2

    The centre, the object and the image lie on one straight line, with the distances from the centre in the ratio of the scale factor.

  10. Identify any invariant points

    invariant point: (1,1)\text{invariant point: } (1, 1)

    The centre of enlargement (1,1)(1, 1) is the one point that stays fixed — every other point moves along a ray through it.

  11. Note a second method

    B′=f(B)⇒B=f−1(B′)B' = f(B) \Rightarrow B = f^{-1}(B')

    Instead of naming the inverse you can write the forward rule with unknowns xx and yy, set it equal to the image point, and solve the two equations. The answer is the same.

  12. Note the mistake to avoid

    k(P−C)≠kPk(P - C) \ne kP

    The scale factor multiplies the vector FROM THE CENTRE to the point, not the coordinates of the point themselves.

  13. Sense check the position of the original point

    ∣k∣=2|k| = 2

    The image is larger than the object and lies on the opposite side of the centre and upside down.

  14. Summarise the method

    image→inverse rule→object\text{image} \rightarrow \text{inverse rule} \rightarrow \text{object}

    To reverse a transformation, apply the inverse transformation to the image: the reverse vector, the same mirror, the opposite turn, or the reciprocal scale factor.

  15. State the coordinates of the original point

    B=(3,2)B = (3, 2)

    The original point is (3, 2)\left(3,\ 2\right).

Answer
B=(3,2)B = (3, 2)
Question 2
5 markschallenging
Triangle ABCABC is rotated 90∘90^{\circ} clockwise about (2,1)(2, 1) to give triangle A′B′C′A'B'C'. The image of vertex AA is the point A′(5,0)A'(5, 0). Find the coordinates of the original point AA.
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Worked solution

  1. Identify the transformation

    rotation,90∘ clockwise,(2,1)\text{rotation}, \quad 90^{\circ} \text{ clockwise}, \quad (2, 1)

    The question describes a rotation of 90 degrees clockwise about (2, 1), but this time the IMAGE is known and the original point is not.

  2. Write down the rule for the transformation

    (u, v)=(x−2, y−1)↦(v, −u)↦(x′, y′)=(2+u′, 1+v′)(u,\ v) = (x - 2,\ y - 1) \mapsto (v,\ -u) \mapsto (x',\ y') = (2 + u',\ 1 + v')

    This rule sends the original point to the image. To go backwards it has to be undone.

  3. Say what undoing the transformation means

    A↦A′(5,0)⇒A′↦AA \mapsto A'(5, 0) \Rightarrow A' \mapsto A

    Reversing a transformation means applying the transformation that takes the image back to the object.

  4. Write down the inverse transformation

    rotation,90∘ anticlockwise,(2,1)\text{rotation}, \quad 90^{\circ} \text{ anticlockwise}, \quad (2, 1)

    The inverse of a rotation of 90 degrees clockwise about (2, 1) is a rotation of 90 degrees anticlockwise about (2, 1).

  5. Write down the rule for the inverse

    (u, v)=(x−2, y−1)↦(−v, u)↦(x′, y′)=(2+u′, 1+v′)(u,\ v) = (x - 2,\ y - 1) \mapsto (-v,\ u) \mapsto (x',\ y') = (2 + u',\ 1 + v')

    This is the rule that will be applied to the image point.

  6. Substitute the image point into the inverse rule

    A′:(3, −1)↦(1, 3)⇒A′′(3,4)A': (3,\ -1) \mapsto (1,\ 3) \Rightarrow A''(3, 4)

    Putting A′(5,0)A'(5, 0) into the inverse rule gives the original point.

  7. Write down the original point

    A=(3,4)A = (3, 4)

    So the vertex that was transformed is A(3,4)A(3, 4).

  8. Check by applying the original transformation

    A:(1, 3)↦(3, −1)⇒A′(5,0)A: (1,\ 3) \mapsto (3,\ -1) \Rightarrow A'(5, 0)

    Transforming A(3,4)A(3, 4) forwards gives (5,0)(5, 0), which is the image the question gave — the answer is right.

  9. Check the distance from the key line or point

    distance2 from centre=10 for both points\text{distance}^2 \text{ from centre} = 10 \text{ for both points}

    Both the object and its image are the same distance from the centre (2,1)(2, 1) — a rotation never changes that distance.

  10. Identify any invariant points

    invariant point: (2,1)\text{invariant point: } (2, 1)

    The only point that does not move under a rotation is the centre itself, (2,1)(2, 1).

  11. Note a second method

    A′=f(A)⇒A=f−1(A′)A' = f(A) \Rightarrow A = f^{-1}(A')

    Instead of naming the inverse you can write the forward rule with unknowns xx and yy, set it equal to the image point, and solve the two equations. The answer is the same.

  12. Note the mistake to avoid

    90∘ clockwise≠90∘ anticlockwise90^{\circ} \text{ clockwise} \ne 90^{\circ} \text{ anticlockwise}

    Turning the wrong way sends the image to a completely different place, so always check the direction stated in the question.

  13. Sense check the position of the original point

    turn: 90∘\text{turn: } 90^{\circ}

    Each vertex stays the same distance from the centre, so the object and image are congruent — only the direction has changed.

  14. Summarise the method

    image→inverse rule→object\text{image} \rightarrow \text{inverse rule} \rightarrow \text{object}

    To reverse a transformation, apply the inverse transformation to the image: the reverse vector, the same mirror, the opposite turn, or the reciprocal scale factor.

  15. State the coordinates of the original point

    A=(3,4)A = (3, 4)

    The original point is (3, 4)\left(3,\ 4\right).

Answer
A=(3,4)A = (3, 4)
Question 3
5 markschallenging
Triangle ABCABC has vertices A(2,3)A(2, 3), B(4,3)B(4, 3) and C(2,4)C(2, 4). Triangle ABCABC is enlarged by scale factor −1-1 to give triangle A′B′C′A'B'C'. Triangle A′B′C′A'B'C' has vertices A′(0,1)A'(0, 1), B′(−2,1)B'(-2, 1) and C′(0,0)C'(0, 0). Find the coordinates of the centre of enlargement.
Show worked solution

Worked solution

  1. Write the object and the image side by side

    A(2,3), B(4,3), C(2,4)→A′(0,1), B′(−2,1), C′(0,0)A(2, 3),\ B(4, 3),\ C(2, 4) \rightarrow A'(0, 1),\ B'(-2, 1),\ C'(0, 0)

    The scale factor is given, so the only thing missing from a full description is the centre.

  2. Write the rule an enlargement obeys

    P′=O+(−1)(P−O)P' = O + \left(-1\right)(P - O)

    Every point PP maps to P′P' so that the vector from the centre OO is multiplied by −1-1. That single equation fixes OO.

  3. Rearrange the rule to make the centre the subject

    O=P′−(−1)P1−(−1)O = \frac{P' - \left(-1\right)P}{1 - \left(-1\right)}

    Expanding gives P′=O+(−1)P−(−1)OP' = O + \left(-1\right)P - \left(-1\right)O, so P′−(−1)P=(1−(−1))OP'-\left(-1\right)P = (1-\left(-1\right))O and the centre can be read off.

  4. Substitute the coordinates of A and its image

    O=12(0−(−1)(2)1−(−1)(3))O = \frac{1}{2}\begin{pmatrix} 0 - (-1)(2) \\ 1 - (-1)(3) \end{pmatrix}

    Using the pair A(2,3)A(2, 3) and A′(0,1)A'(0, 1) turns the rule into a calculation with no unknowns left except the centre.

  5. Work out the centre

    O=(1,2)O = (1, 2)

    The centre of enlargement is (1,2)(1, 2).

  6. Check the centre with vertex B

    (1,2)+(−1)(4−13−2)=(−2,1)(1, 2) + \left(-1\right)\begin{pmatrix} 4 - 1 \\ 3 - 2 \end{pmatrix} = (-2, 1)

    Using the same centre and scale factor on B(4,3)B(4, 3) gives B′(−2,1)B'(-2, 1), exactly as required.

  7. Check the centre with vertex C

    (1,2)+(−1)(2−14−2)=(0,0)(1, 2) + \left(-1\right)\begin{pmatrix} 2 - 1 \\ 4 - 2 \end{pmatrix} = (0, 0)

    The third vertex works too, so (1,2)(1, 2) is the centre for the whole triangle and not just for one vertex.

  8. Check by drawing the rays through matching vertices

    AA′, BB′, CC′ meet at (1,2)AA', \ BB', \ CC' \text{ meet at } (1, 2)

    Joining each vertex to its image and extending the lines, all three rays cross at (1,2)(1, 2) — the standard construction for the centre.

  9. Check the sign of the scale factor against the picture

    k=−1k = -1

    A negative scale factor puts the image on the opposite side of the centre and turns it upside down, which is what the picture shows.

  10. Check the invariant point

    (1,2)↦(1,2)(1, 2) \mapsto (1, 2)

    The centre is the one point that does not move: substituting (1,2)(1, 2) into the rule returns (1,2)(1, 2).

  11. Note a second method

    O=P′−(−1)P1−(−1) for any vertex PO = \frac{P' - \left(-1\right)P}{1 - \left(-1\right)} \text{ for any vertex } P

    Any one pair of matching vertices gives the centre; the other two pairs are then a free check. Drawing the rays is the same idea done with a ruler.

  12. Note the mistake to avoid

    O≠midpoint of AA′O \ne \text{midpoint of } AA'

    The centre is not the midpoint of AA′AA' — it is the point where all the rays meet, and it usually lies outside the segment.

  13. Summarise the method

    one vertex pair+k→O\text{one vertex pair} + k \rightarrow O

    Substitute one matching pair of vertices into the enlargement rule, rearrange for the centre, then check the answer on the other vertices.

  14. Sense check the position of the centre

    (1,2)(1, 2)

    The centre lies on every ray through a vertex and its image, so it must sit where those rays cross — read the picture and the algebra together.

  15. State the centre of enlargement

    O=(1,2)O = (1, 2)

    The centre of enlargement is (1,2)(1, 2).

Answer
O=(1,2)O = (1, 2)
Question 4
6 markschallenging
Triangle ABCABC has vertices A(3,2)A(3, 2), B(5,2)B(5, 2) and C(3,3)C(3, 3). Triangle ABCABC is enlarged by scale factor −2-2 to give triangle A′B′C′A'B'C'. Triangle A′B′C′A'B'C' has vertices A′(−3,−1)A'(-3, -1), B′(−7,−1)B'(-7, -1) and C′(−3,−3)C'(-3, -3). Find the coordinates of the centre of enlargement.
Show worked solution

Worked solution

  1. Write the object and the image side by side

    A(3,2), B(5,2), C(3,3)→A′(−3,−1), B′(−7,−1), C′(−3,−3)A(3, 2),\ B(5, 2),\ C(3, 3) \rightarrow A'(-3, -1),\ B'(-7, -1),\ C'(-3, -3)

    The scale factor is given, so the only thing missing from a full description is the centre.

  2. Write the rule an enlargement obeys

    P′=O+(−2)(P−O)P' = O + \left(-2\right)(P - O)

    Every point PP maps to P′P' so that the vector from the centre OO is multiplied by −2-2. That single equation fixes OO.

  3. Rearrange the rule to make the centre the subject

    O=P′−(−2)P1−(−2)O = \frac{P' - \left(-2\right)P}{1 - \left(-2\right)}

    Expanding gives P′=O+(−2)P−(−2)OP' = O + \left(-2\right)P - \left(-2\right)O, so P′−(−2)P=(1−(−2))OP'-\left(-2\right)P = (1-\left(-2\right))O and the centre can be read off.

  4. Substitute the coordinates of A and its image

    O=13(−3−(−2)(3)−1−(−2)(2))O = \frac{1}{3}\begin{pmatrix} -3 - (-2)(3) \\ -1 - (-2)(2) \end{pmatrix}

    Using the pair A(3,2)A(3, 2) and A′(−3,−1)A'(-3, -1) turns the rule into a calculation with no unknowns left except the centre.

  5. Work out the centre

    O=(1,1)O = (1, 1)

    The centre of enlargement is (1,1)(1, 1).

  6. Check the centre with vertex B

    (1,1)+(−2)(5−12−1)=(−7,−1)(1, 1) + \left(-2\right)\begin{pmatrix} 5 - 1 \\ 2 - 1 \end{pmatrix} = (-7, -1)

    Using the same centre and scale factor on B(5,2)B(5, 2) gives B′(−7,−1)B'(-7, -1), exactly as required.

  7. Check the centre with vertex C

    (1,1)+(−2)(3−13−1)=(−3,−3)(1, 1) + \left(-2\right)\begin{pmatrix} 3 - 1 \\ 3 - 1 \end{pmatrix} = (-3, -3)

    The third vertex works too, so (1,1)(1, 1) is the centre for the whole triangle and not just for one vertex.

  8. Check by drawing the rays through matching vertices

    AA′, BB′, CC′ meet at (1,1)AA', \ BB', \ CC' \text{ meet at } (1, 1)

    Joining each vertex to its image and extending the lines, all three rays cross at (1,1)(1, 1) — the standard construction for the centre.

  9. Check the sign of the scale factor against the picture

    k=−2k = -2

    A negative scale factor puts the image on the opposite side of the centre and turns it upside down, which is what the picture shows.

  10. Check the invariant point

    (1,1)↦(1,1)(1, 1) \mapsto (1, 1)

    The centre is the one point that does not move: substituting (1,1)(1, 1) into the rule returns (1,1)(1, 1).

  11. Note a second method

    O=P′−(−2)P1−(−2) for any vertex PO = \frac{P' - \left(-2\right)P}{1 - \left(-2\right)} \text{ for any vertex } P

    Any one pair of matching vertices gives the centre; the other two pairs are then a free check. Drawing the rays is the same idea done with a ruler.

  12. Note the mistake to avoid

    O≠midpoint of AA′O \ne \text{midpoint of } AA'

    The centre is not the midpoint of AA′AA' — it is the point where all the rays meet, and it usually lies outside the segment.

  13. Summarise the method

    one vertex pair+k→O\text{one vertex pair} + k \rightarrow O

    Substitute one matching pair of vertices into the enlargement rule, rearrange for the centre, then check the answer on the other vertices.

  14. Sense check the position of the centre

    (1,1)(1, 1)

    The centre lies on every ray through a vertex and its image, so it must sit where those rays cross — read the picture and the algebra together.

  15. State the centre of enlargement

    O=(1,1)O = (1, 1)

    The centre of enlargement is (1,1)(1, 1).

Answer
O=(1,1)O = (1, 1)
Question 5
6 markschallenging
Triangle ABCABC has vertices A(1,1)A(1, 1), B(2,1)B(2, 1) and C(1,3)C(1, 3). Triangle ABCABC is enlarged by scale factor 22, centre (0,0)(0, 0) to give triangle A′B′C′A'B'C'. Triangle A′B′C′A'B'C' is then enlarged by scale factor −12-\frac{1}{2}, centre (0,0)(0, 0) to give triangle A′′B′′C′′A''B''C''. Describe fully the single transformation that maps triangle ABCABC onto triangle A′′B′′C′′A''B''C''.
Show worked solution

Worked solution

  1. Identify the first transformation

    enlargement,k=2,(0,0)\text{enlargement}, \quad k = 2, \quad (0, 0)

    The first instruction is a enlargement with scale factor 2, centre (0, 0).

  2. Write down the rule for the first transformation

    (x, y)↦(2x, 2y)(x,\ y) \mapsto (2x,\ 2y)

    Apply it to every vertex of triangle ABCABC.

  3. Apply the first rule to vertex A

    A:(1, 1)×2=(2, 2)⇒A′(2,2)A: (1,\ 1) \times 2 = (2,\ 2) \Rightarrow A'(2, 2)

    A(1,1)A(1, 1) maps to A′(2,2)A'(2, 2).

  4. Apply the first rule to vertex B

    B:(2, 1)×2=(4, 2)⇒B′(4,2)B: (2,\ 1) \times 2 = (4,\ 2) \Rightarrow B'(4, 2)

    B(2,1)B(2, 1) maps to B′(4,2)B'(4, 2).

  5. Apply the first rule to vertex C

    C:(1, 3)×2=(2, 6)⇒C′(2,6)C: (1,\ 3) \times 2 = (2,\ 6) \Rightarrow C'(2, 6)

    C(1,3)C(1, 3) maps to C′(2,6)C'(2, 6).

  6. Write down the first image triangle

    A′(2,2), B′(4,2), C′(2,6)A'(2, 2),\ B'(4, 2),\ C'(2, 6)

    Triangle A′B′C′A'B'C' has vertices A′(2,2), B′(4,2), C′(2,6)A'(2, 2),\ B'(4, 2),\ C'(2, 6).

  7. Write down the rule for the second transformation

    (x, y)↦(−12x, −12y)(x,\ y) \mapsto (-\frac{1}{2}x,\ -\frac{1}{2}y)

    The second instruction is a enlargement with scale factor -1/2, centre (0, 0), applied to triangle A′B′C′A'B'C'.

  8. Apply the second rule to the image of A

    A′:(2, 2)×(−12)=(−1, −1)⇒A′′(−1,−1)A': (2,\ 2) \times \left(-\frac{1}{2}\right) = (-1,\ -1) \Rightarrow A''(-1, -1)

    A′(2,2)A'(2, 2) maps to A′′(−1,−1)A''(-1, -1).

  9. Apply the second rule to the image of B

    B′:(4, 2)×(−12)=(−2, −1)⇒B′′(−2,−1)B': (4,\ 2) \times \left(-\frac{1}{2}\right) = (-2,\ -1) \Rightarrow B''(-2, -1)

    B′(4,2)B'(4, 2) maps to B′′(−2,−1)B''(-2, -1).

  10. Apply the second rule to the image of C

    C′:(2, 6)×(−12)=(−1, −3)⇒C′′(−1,−3)C': (2,\ 6) \times \left(-\frac{1}{2}\right) = (-1,\ -3) \Rightarrow C''(-1, -3)

    C′(2,6)C'(2, 6) maps to C′′(−1,−3)C''(-1, -3).

  11. Write down the final image triangle

    A′′(−1,−1), B′′(−2,−1), C′′(−1,−3)A''(-1, -1),\ B''(-2, -1),\ C''(-1, -3)

    Triangle A′′B′′C′′A''B''C'' has vertices A′′(−1,−1),B′′(−2,−1),C′′(−1,−3)A''(-1, -1), B''(-2, -1), C''(-1, -3). Now compare it with the original triangle ABCABC.

  12. Work out the single transformation that maps the object to the final image

    k=A′B′AB=−1,centre (0,0)k = \frac{A'B'}{AB} = -1, \quad \text{centre } (0, 0)

    Matching lengths are in the ratio 11, and the image is on the opposite side of the centre, so the scale factor is negative: k=−1k = -1. The lines through matching vertices all meet at (0,0)(0, 0).

  13. Rule out alternative number 1

    enlargement,k=1,(0,0):A(1,1)↦(1, 1)≠A′′(−1,−1)\text{enlargement}, \quad k = 1, \quad (0, 0): A(1, 1) \mapsto \left(1,\ 1\right) \ne A''(-1, -1)

    Enlargement with scale factor 11, centre (0,0)(0, 0) would send A(1,1)A(1, 1) to (1, 1)\left(1,\ 1\right), but the final image of AA is A′′(−1,−1)A''(-1, -1) — so it is not the right description.

  14. Rule out alternative number 2

    enlargement,k=−2,(0,0):A(1,1)↦(−2, −2)≠A′′(−1,−1)\text{enlargement}, \quad k = -2, \quad (0, 0): A(1, 1) \mapsto \left(-2,\ -2\right) \ne A''(-1, -1)

    Enlargement with scale factor −2-2, centre (0,0)(0, 0) would send A(1,1)A(1, 1) to (−2, −2)\left(-2,\ -2\right), but the final image of AA is A′′(−1,−1)A''(-1, -1) — so it is not the right description.

  15. State the single transformation

    Enlargement with scale factor -1, centre (0, 0)\text{Enlargement with scale factor -1, centre (0, 0)}

    Enlargement with scale factor −1-1, centre (0,0)(0, 0) maps triangle ABCABC straight onto triangle A′′B′′C′′A''B''C'', so it is the single transformation with the same effect as the two together.

Answer
Enlargement with scale factor -1, centre (0, 0)\text{Enlargement with scale factor -1, centre (0, 0)}

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