Transformations Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE Transformations questions. See exactly how to solve problems on translation, column vector, negative vector, reflection.

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.

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(5,1)B(5, 1) and C(2,4)C(2, 4). Triangle ABCABC is translated by the column vector (−42)\begin{pmatrix} -4 \\ 2 \end{pmatrix}. Find the coordinates of the image of BB.

Worked solution

  1. Write down the coordinate rule for the transformation

    (x, y)↦(x−4, y+2)(x,\ y) \mapsto (x - 4,\ y + 2)

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

  2. Apply the rule to vertex B

    B(5,1)↦(5−4, 1+2)=B′(1,3)B(5, 1) \mapsto (5 - 4,\ 1 + 2) = B'(1, 3)

    The same rule sends B(5,1)B(5, 1) to B′(1,3)B'(1, 3).

  3. State the coordinates asked for

    B′=(1,3)B' = (1, 3)

    So the image of BB is (1, 3)\left(1,\ 3\right).

Answer
B′=(1,3)B' = (1, 3)
Question 3
2 markseasy
Triangle ABCABC has vertices A(−1,2)A(-1, 2), B(2,2)B(2, 2) and C(−1,5)C(-1, 5). Triangle ABCABC is translated by the column vector (3−4)\begin{pmatrix} 3 \\ -4 \end{pmatrix}. Find the coordinates of the image of CC.

Worked solution

  1. Write down the coordinate rule for the transformation

    (x, y)↦(x+3, y−4)(x,\ y) \mapsto (x + 3,\ y - 4)

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

  2. Apply the rule to vertex C

    C(−1,5)↦(−1+3, 5−4)=C′(2,1)C(-1, 5) \mapsto (-1 + 3,\ 5 - 4) = C'(2, 1)

    And C(−1,5)C(-1, 5) maps to C′(2,1)C'(2, 1).

  3. State the coordinates asked for

    C′=(2,1)C' = (2, 1)

    So the image of CC is (2, 1)\left(2,\ 1\right).

Answer
C′=(2,1)C' = (2, 1)
Question 4
1 markeasy
Triangle ABCABC has vertices A(0,0)A(0, 0), B(3,0)B(3, 0) and C(0,2)C(0, 2). Triangle ABCABC is translated by the column vector (−2−3)\begin{pmatrix} -2 \\ -3 \end{pmatrix}. Find the coordinates of the image of BB.

Worked solution

  1. Write down the coordinate rule for the transformation

    (x, y)↦(x−2, y−3)(x,\ y) \mapsto (x - 2,\ y - 3)

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

  2. Apply the rule to vertex B

    B(3,0)↦(3−2, 0−3)=B′(1,−3)B(3, 0) \mapsto (3 - 2,\ 0 - 3) = B'(1, -3)

    The same rule sends B(3,0)B(3, 0) to B′(1,−3)B'(1, -3).

  3. State the coordinates asked for

    B′=(1,−3)B' = (1, -3)

    So the image of BB is (1, −3)\left(1,\ -3\right).

Answer
B′=(1,−3)B' = (1, -3)
Question 5
2 markseasy
Triangle ABCABC has vertices A(1,1)A(1, 1), B(3,1)B(3, 1) and C(1,4)C(1, 4). Triangle ABCABC is reflected in the line x=4x = 4. Find the coordinates of the image of AA.

Worked solution

  1. Write down the coordinate rule for the transformation

    (x, y)↦(8−x, y)(x,\ y) \mapsto (8 - x,\ y)

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

  2. Apply the rule to vertex A

    A(1,1)↦(8−1, 1)=A′(7,1)A(1, 1) \mapsto (8 - 1,\ 1) = A'(7, 1)

    Substituting A(1,1)A(1, 1) into the rule gives the image point A′(7,1)A'(7, 1).

  3. State the coordinates asked for

    A′=(7,1)A' = (7, 1)

    So the image of AA is (7, 1)\left(7,\ 1\right).

Answer
A′=(7,1)A' = (7, 1)

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