GCSE Combining transformations Practice Questions

Apply the first transformation, the reflection in the line $x = 0$, to the point.

The rule for a reflection in the line $x = 0$ is $(x,\, y) \rightarrow (-x,\, y)$, so $A(3, 1)$ moves to $(-3, 1)$.

Apply the second transformation, the translation by the vector (2, 3), to that image.

The second transformation acts on the image $(-3, 1)$, not on the original point. Its rule is $(x,\, y) \rightarrow (x + 2,\, y + 3)$.

State the coordinates of the final image.

The point ends at $(-1, 4)$. Doing the two transformations in the other order would generally land somewhere else.

Apply the first transformation to every vertex.

The rule for a reflection in the line $x = 2$ is $(x,\, y) \rightarrow (4 - x,\, y)$, which gives $A'(1,\, 1),\; B'(-1,\, 1),\; C'(-1,\, 4)$.

Apply the second transformation to every vertex of that image.

The rule for a reflection in the line $x = -1$ is $(x,\, y) \rightarrow (-x - 2,\, y)$, which gives triangle $U$: $A'''(-3,\, 1),\; B'''(-1,\, 1),\; C'''(-1,\, 4)$.

Describe the single transformation that does the same job.

The single transformation equivalent to a reflection in the line $x = 2$ followed by a reflection in the line $x = -1$ is: a translation by the vector $\begin{pmatrix} -6 \\ 0 \end{pmatrix}$. It has been checked on every vertex.

Write down the vertices of triangle T and the two transformations.

Triangle $T$ has vertices $A(3,\, 1),\; B(5,\, 1),\; C(5,\, 4)$. The two transformations are a translation by the vector $\begin{pmatrix} -2 \\ 1 \end{pmatrix}$ and a reflection in the line $y = -x$.

First order: map every vertex through the first transformation, then the second.

Using $(x,\, y) \rightarrow (x - 2,\, y + 1)$ and then $(x,\, y) \rightarrow (-y,\, -x)$ gives triangle $P$: $A_{P}(-2,\, -1),\; B_{P}(-2,\, -3),\; C_{P}(-5,\, -3)$.

Second order: map every vertex through the second transformation, then the first.

Using $(x,\, y) \rightarrow (-y,\, -x)$ and then $(x,\, y) \rightarrow (x - 2,\, y + 1)$ gives triangle $Q$: $A_{Q}(-3,\, -2),\; B_{Q}(-3,\, -4),\; C_{Q}(-6,\, -4)$.

Compare triangle P with triangle Q.

$P$ and $Q$ are in different positions, so the order of the two transformations matters.

Pick out the vertex the question asks for.

Triangle $P$ is the one built in that order, so the image of $B$ is $(-2, -3)$ (not $(-3, -4)$).

State the final answer.

On triangle $P$ the image of $B$ is $(-2, -3)$. In the other order it is $(-3, -4)$ — a different point, which is why the order of the transformations must be respected.

Write down the coordinates of the vertices of the original triangle.

Triangle $T$ has vertices $A(2,\, 4),\; B(6,\, 4),\; C(2,\, 8)$. Every vertex must be mapped — checking one is not enough.

Write the coordinate rule for the first transformation, the enlargement with scale factor 1/2 about (0, 0).

For an enlargement with scale factor $\frac{1}{2}$, centre $(0, 0)$ the rule is $(x,\, y) \rightarrow (\frac{1}{2}x,\, \frac{1}{2}y)$.

Apply the first transformation to every vertex.

Using $(x,\, y) \rightarrow (\frac{1}{2}x,\, \frac{1}{2}y)$ on each vertex gives $A'(1,\, 2),\; B'(3,\, 2),\; C'(1,\, 4)$.

Write the coordinate rule for the second transformation, the enlargement with scale factor -4 about (2, 2).

For an enlargement with scale factor $-4$, centre $(2, 2)$ the rule is $(x,\, y) \rightarrow (10 - 4x,\, 10 - 4y)$.

Apply the second transformation to every vertex of that image.

Using $(x,\, y) \rightarrow (10 - 4x,\, 10 - 4y)$ on each vertex gives triangle $U$ with vertices $A'''(6,\, 2),\; B'''(-2,\, 2),\; C'''(6,\, -6)$.

Decide what kind of single transformation maps T onto U.

Comparing $T$ with $U$: every length is multiplied by $2$, so the combination is an enlargement.

Find the parameters of that single transformation.

The centre is the only invariant point, $(\frac{10}{3}, \frac{10}{3})$, and the scale factor is $-2$.

Check the candidate transformation on all three vertices.

Enlargement with scale factor $-2$, centre $(\frac{10}{3}, \frac{10}{3})$ sends every vertex of $T$ to the matching vertex of $U$, so it really is equivalent to the pair of transformations.

Check the answer by a second route.

Enlarging by $\frac{1}{2}$ and then by $-4$ multiplies every length by $\frac{1}{2} \times -4 = -2$.

State the answer.

The two scale factors multiply: the single enlargement has scale factor $-2$.

Write down the coordinates of the vertices of the original triangle.

Triangle $T$ has vertices $A(2,\, 1),\; B(5,\, 1),\; C(2,\, 3)$. Every vertex must be mapped — checking one is not enough.

Write the coordinate rule for the first transformation, the rotation of 180 degrees about (1, 1).

For a rotation of $180^{\circ}$ about the point $(1, 1)$ the rule is $(x,\, y) \rightarrow (2 - x,\, 2 - y)$.

Apply the first transformation to vertex A.

Substituting $(2, 1)$ into the rule gives $(0, 1)$.

Apply the first transformation to vertex B.

Substituting $(5, 1)$ into the rule gives $(-3, 1)$.

Apply the first transformation to vertex C.

Substituting $(2, 3)$ into the rule gives $(0, -1)$.

State the intermediate image after the first transformation.

The intermediate triangle has vertices $A'(0,\, 1),\; B'(-3,\, 1),\; C'(0,\, -1)$. The second transformation is applied to this triangle.

Write the coordinate rule for the second transformation, the rotation of 90 degrees clockwise about (3, 1).

For a rotation of $90^{\circ}$ clockwise about the point $(3, 1)$ the rule is $(x,\, y) \rightarrow (y + 2,\, 4 - x)$.

Apply the second transformation to the image of A.

$(0, 1)$ maps to $(3, 4)$.

Apply the second transformation to the image of B.

$(-3, 1)$ maps to $(3, 7)$.

Apply the second transformation to the image of C.

$(0, -1)$ maps to $(1, 4)$.

State the final image, triangle U.

Triangle $U$ has vertices $A'''(3,\, 4),\; B'''(3,\, 7),\; C'''(1,\, 4)$.

Decide what kind of single transformation maps T onto U.

Comparing $T$ with $U$: lengths are unchanged and the sense of the lettering is unchanged, so the combination is a rotation.

Find the parameters of that single transformation.

The centre is the only point that does not move: $(1, 3)$.

Check the candidate transformation on all three vertices.

Rotation of $90^{\circ}$ anticlockwise about the point $(1, 3)$ sends every vertex of $T$ to the matching vertex of $U$, so it really is equivalent to the pair of transformations.

State the answer.

The centre of the rotation is $(1, 3)$ — the one point that both transformations together leave exactly where it started.