GCSE Transformations Practice Questions

Free GCSE Transformations practice questions with full step-by-step worked solutions. Covers translation, column vector, negative vector, reflection. Practise exam-style problems and check your method.

translationcolumn vectornegative vectorreflectionvertical mirror linehorizontal mirror line
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
Triangle ABCABC has vertices A(1,1)A(1, 1), B(4,1)B(4, 1) and C(1,3)C(1, 3). Triangle ABCABC is translated by the column vector (32)\begin{pmatrix} 3 \\ 2 \end{pmatrix}. Find the coordinates of the image of AA.
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Worked solution

  1. Write down the coordinate rule for the transformation

    (x, y)(x+3, y+2)(x,\ y) \mapsto (x + 3,\ y + 2)

    This rule turns the words into arithmetic: apply it to each vertex in turn.

  2. Apply the rule to vertex A

    A(1,1)(1+3, 1+2)=A(4,3)A(1, 1) \mapsto (1 + 3,\ 1 + 2) = A'(4, 3)

    Substituting A(1,1)A(1, 1) into the rule gives the image point A(4,3)A'(4, 3).

  3. State the coordinates asked for

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

    So the image of AA is (4, 3)\left(4,\ 3\right).

Answer
A=(4,3)A' = (4, 3)
Question 2
2 markseasy
Triangle ABCABC has vertices A(2,1)A(-2, 1), B(1,1)B(1, 1) and C(2,3)C(-2, 3). Triangle ABCA'B'C' has vertices A(2,1)A'(2, -1), B(5,1)B'(5, -1) and C(2,1)C'(2, 1). Describe fully the single transformation that maps triangle ABCABC onto triangle ABCA'B'C'.
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Worked solution

  1. Write the object and the image side by side

    A(2,1), B(1,1), C(2,3)A(2,1), B(5,1), C(2,1)A(-2, 1),\ B(1, 1),\ C(-2, 3) \rightarrow A'(2, -1),\ B'(5, -1),\ C'(2, 1)

    Matching vertices are AAA \rightarrow A', BBB \rightarrow B' and CCC \rightarrow C'; the whole description has to be read from those three pairs.

  2. Find the information that pins the transformation down

    (42)=(2(2)11)\begin{pmatrix} 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 2 - (-2) \\ -1 - 1 \end{pmatrix}

    Subtracting the object vertex from its image gives the column vector (42)\begin{pmatrix} 4 \\ -2 \end{pmatrix}, and every other vertex moves by the same vector.

  3. State the full description

    Translation by the column vector (4, -2)\text{Translation by the column vector (4, -2)}

    Translation by the column vector (42)\begin{pmatrix} 4 \\ -2 \end{pmatrix} is the single transformation that maps triangle ABCABC onto triangle ABCA'B'C'.

Answer
Translation by the column vector (4, -2)\text{Translation by the column vector (4, -2)}
Question 3
2 marksintermediate
Triangle ABCABC has vertices A(2,1)A(2, 1), B(5,1)B(5, 1) and C(2,2)C(2, 2). Triangle ABCA'B'C' has vertices A(2,5)A'(2, 5), B(5,5)B'(5, 5) and C(2,4)C'(2, 4). Describe fully the single transformation that maps triangle ABCABC onto triangle ABCA'B'C'.
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Worked solution

  1. Write the object and the image side by side

    A(2,1), B(5,1), C(2,2)A(2,5), B(5,5), C(2,4)A(2, 1),\ B(5, 1),\ C(2, 2) \rightarrow A'(2, 5),\ B'(5, 5),\ C'(2, 4)

    Matching vertices are AAA \rightarrow A', BBB \rightarrow B' and CCC \rightarrow C'; the whole description has to be read from those three pairs.

  2. Decide which type of transformation it is

    type: reflect\text{type: } \text{reflect}

    The shape is congruent but flipped over, so this is a reflection; the mirror line is what is left to find.

  3. Find the information that pins the transformation down

    y=1+52=3y = \frac{1 + 5}{2} = 3

    AA and AA' have the same xx-coordinate, and the mirror is the horizontal line halfway between them: y=3y = 3.

  4. Check the description maps every vertex correctly

    AA(2,5), BB(5,5), CC(2,4)A \mapsto A'(2, 5), \ B \mapsto B'(5, 5), \ C \mapsto C'(2, 4)

    The reflection in the line y=3y = 3 sends all three vertices to the right places, so the description is complete and correct.

  5. Test alternative number 1

    reflection,y=2:A(2,1)(2, 3)A(2,5)\text{reflection}, \quad y = 2: A(2, 1) \mapsto \left(2,\ 3\right) \ne A'(2, 5)

    Reflection in the line y=2y = 2 would send A(2,1)A(2, 1) to (2, 3)\left(2,\ 3\right), but the image of AA is A(2,5)A'(2, 5) — so that description is wrong.

  6. State the full description

    Reflection in the line y = 3\text{Reflection in the line y = 3}

    Reflection in the line y=3y = 3 is the single transformation that maps triangle ABCABC onto triangle ABCA'B'C'.

Answer
Reflection in the line y = 3\text{Reflection in the line y = 3}
Question 4
4 markshard
Triangle ABCABC has vertices A(2,2)A(2, 2), B(4,2)B(4, 2) and C(2,3)C(2, 3). Triangle ABCA'B'C' has vertices A(3,3)A'(3, 3), B(7,3)B'(7, 3) and C(3,5)C'(3, 5). Describe fully the single transformation that maps triangle ABCABC onto triangle ABCA'B'C'.
Show worked solution

Worked solution

  1. Write the object and the image side by side

    A(2,2), B(4,2), C(2,3)A(3,3), B(7,3), C(3,5)A(2, 2),\ B(4, 2),\ C(2, 3) \rightarrow A'(3, 3),\ B'(7, 3),\ C'(3, 5)

    Matching vertices are AAA \rightarrow A', BBB \rightarrow B' and CCC \rightarrow C'; the whole description has to be read from those three pairs.

  2. Compare a pair of matching side lengths

    AB2=4,AB2=16AB^2 = 4, \quad A'B'^2 = 16

    The image side is longer or shorter than the object side, so the shapes are not congruent — this has to be an enlargement.

  3. Decide which type of transformation it is

    type: enlarge\text{type: } \text{enlarge}

    The shape has changed size, so this is an enlargement; the scale factor and the centre are what is left to find.

  4. Find the information that pins the transformation down

    k=ABAB=2,centre (1,1)k = \frac{A'B'}{AB} = 2, \quad \text{centre } (1, 1)

    Matching lengths are in the ratio 22, so k=2k = 2. The lines through matching vertices all meet at (1,1)(1, 1).

  5. Check the description maps every vertex correctly

    AA(3,3), BB(7,3), CC(3,5)A \mapsto A'(3, 3), \ B \mapsto B'(7, 3), \ C \mapsto C'(3, 5)

    The enlargement with scale factor 2, centre (1, 1) sends all three vertices to the right places, so the description is complete and correct.

  6. Test alternative number 1

    enlargement,k=2,(0,0):A(2,2)(4, 4)A(3,3)\text{enlargement}, \quad k = 2, \quad (0, 0): A(2, 2) \mapsto \left(4,\ 4\right) \ne A'(3, 3)

    Enlargement with scale factor 22, centre (0,0)(0, 0) would send A(2,2)A(2, 2) to (4, 4)\left(4,\ 4\right), but the image of AA is A(3,3)A'(3, 3) — so that description is wrong.

  7. Test alternative number 2

    enlargement,k=3,(1,1):A(2,2)(4, 4)A(3,3)\text{enlargement}, \quad k = 3, \quad (1, 1): A(2, 2) \mapsto \left(4,\ 4\right) \ne A'(3, 3)

    Enlargement with scale factor 33, centre (1,1)(1, 1) would send A(2,2)A(2, 2) to (4, 4)\left(4,\ 4\right), but the image of AA is A(3,3)A'(3, 3) — so that description is wrong.

  8. Test alternative number 3

    enlargement,k=2,(1,1):A(2,2)(1, 1)A(3,3)\text{enlargement}, \quad k = -2, \quad (1, 1): A(2, 2) \mapsto \left(-1,\ -1\right) \ne A'(3, 3)

    Enlargement with scale factor 2-2, centre (1,1)(1, 1) would send A(2,2)A(2, 2) to (1, 1)\left(-1,\ -1\right), but the image of AA is A(3,3)A'(3, 3) — so that description is wrong.

  9. Test alternative number 4

    enlargement,k=12,(1,1):A(2,2)(32, 32)A(3,3)\text{enlargement}, \quad k = \frac{1}{2}, \quad (1, 1): A(2, 2) \mapsto \left(\frac{3}{2},\ \frac{3}{2}\right) \ne A'(3, 3)

    Enlargement with scale factor 12\frac{1}{2}, centre (1,1)(1, 1) would send A(2,2)A(2, 2) to (32, 32)\left(\frac{3}{2},\ \frac{3}{2}\right), but the image of AA is A(3,3)A'(3, 3) — so that description is wrong.

  10. State the full description

    Enlargement with scale factor 2, centre (1, 1)\text{Enlargement with scale factor 2, centre (1, 1)}

    Enlargement with scale factor 22, centre (1,1)(1, 1) is the single transformation that maps triangle ABCABC onto triangle ABCA'B'C'.

Answer
Enlargement with scale factor 2, centre (1, 1)\text{Enlargement with scale factor 2, centre (1, 1)}
Question 5
6 markschallenging
Triangle ABCABC is enlarged by scale factor 2-2, centre (1,1)(1, 1) to give triangle ABCA'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)(12(x1), 12(y1))(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

    BB(3,1)BBB \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)(112(x1), 112(y1))(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=f1(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(PC)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

    imageinverse ruleobject\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)

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