Hard GCSE Combining transformations Questions

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.

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

Triangle $T$ has vertices $A(1,\, -3),\; B(4,\, -3),\; C(4,\, -1)$. Every vertex must be mapped — checking one is not enough.

Write the coordinate rule for the first transformation, the enlargement with scale factor 3 about (1, 1).

For an enlargement with scale factor $3$, centre $(1, 1)$ the rule is $(x,\, y) \rightarrow (3x - 2,\, 3y - 2)$.

Apply the first transformation to vertex A.

Substituting $(1, -3)$ into the rule gives $(1, -11)$.

Apply the first transformation to vertex B.

Substituting $(4, -3)$ into the rule gives $(10, -11)$.

Apply the first transformation to vertex C.

Substituting $(4, -1)$ into the rule gives $(10, -5)$.

State the intermediate image after the first transformation.

The intermediate triangle has vertices $A'(1,\, -11),\; B'(10,\, -11),\; C'(10,\, -5)$. The second transformation is applied to this triangle.

Write the coordinate rule for the second transformation, the translation by the vector (-4, 2).

For a translation by the vector $\begin{pmatrix} -4 \\ 2 \end{pmatrix}$ the rule is $(x,\, y) \rightarrow (x - 4,\, y + 2)$.

Apply the second transformation to the image of A.

$(1, -11)$ maps to $(-3, -9)$.

Apply the second transformation to the image of B.

$(10, -11)$ maps to $(6, -9)$.

Apply the second transformation to the image of C.

$(10, -5)$ maps to $(6, -3)$.

State the final image, triangle U.

Triangle $U$ has vertices $A'''(-3,\, -9),\; B'''(6,\, -9),\; C'''(6,\, -3)$.

Decide what kind of single transformation maps T onto U.

Comparing $T$ with $U$: every length is multiplied by $3$, so the combination is an enlargement.

Find the parameters of that single transformation.

The centre is the only invariant point, $(3, 0)$, and the scale factor is $3$.

Check the candidate transformation on all three vertices.

Enlargement with scale factor $3$, centre $(3, 0)$ 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 enlargement is $(3, 0)$ — the one point that both transformations together leave exactly where it started.

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

Triangle $T$ has vertices $A(1,\, 1),\; B(4,\, 1),\; C(4,\, 3)$. Every vertex must be mapped — checking one is not enough.

Write the coordinate rule for the first transformation, the rotation of 90 degrees anticlockwise about (1, 1).

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

Apply the first transformation to vertex A.

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

Apply the first transformation to vertex B.

Substituting $(4, 1)$ into the rule gives $(1, 4)$.

Apply the first transformation to vertex C.

Substituting $(4, 3)$ into the rule gives $(-1, 4)$.

State the intermediate image after the first transformation.

The intermediate triangle has vertices $A'(1,\, 1),\; B'(1,\, 4),\; C'(-1,\, 4)$. The second transformation is applied to this triangle.

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

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

Apply the second transformation to the image of A.

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

Apply the second transformation to the image of B.

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

Apply the second transformation to the image of C.

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

State the final image, triangle U.

Triangle $U$ has vertices $A'''(3,\, 3),\; B'''(0,\, 3),\; C'''(0,\, 1)$.

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: $(2, 2)$.

Check the candidate transformation on all three vertices.

Rotation of $180^{\circ}$ about the point $(2, 2)$ 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 $(2, 2)$ — the one point that both transformations together leave exactly where it started.

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

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

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

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

Apply the first transformation to vertex A.

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

Apply the first transformation to vertex B.

Substituting $(6, 2)$ into the rule gives $(11, 3)$.

Apply the first transformation to vertex C.

Substituting $(6, 4)$ into the rule gives $(11, 7)$.

State the intermediate image after the first transformation.

The intermediate triangle has vertices $A'(3,\, 3),\; B'(11,\, 3),\; C'(11,\, 7)$. The second transformation is applied to this triangle.

Write the coordinate rule for the second transformation, the translation by the vector (2, -6).

For a translation by the vector $\begin{pmatrix} 2 \\ -6 \end{pmatrix}$ the rule is $(x,\, y) \rightarrow (x + 2,\, y - 6)$.

Apply the second transformation to the image of A.

$(3, 3)$ maps to $(5, -3)$.

Apply the second transformation to the image of B.

$(11, 3)$ maps to $(13, -3)$.

Apply the second transformation to the image of C.

$(11, 7)$ maps to $(13, 1)$.

State the final image, triangle U.

Triangle $U$ has vertices $A'''(5,\, -3),\; B'''(13,\, -3),\; C'''(13,\, 1)$.

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.

Reverse that single transformation.

The forward transformation is an enlargement with scale factor $2$, centre $(-1, 7)$. Undoing it gives an enlargement with scale factor $\frac{1}{2}$, centre $(-1, 7)$. The centre is the only invariant point, $(-1, 7)$, and the scale factor is $\frac{1}{2}$.

Check the reversed transformation maps U back onto T.

Enlargement with scale factor $\frac{1}{2}$, centre $(-1, 7)$ sends every vertex of $U$ back to the matching vertex of $T$.

State the answer.

Going forwards, $T$ maps onto $U$ by an enlargement with scale factor $2$, centre $(-1, 7)$. Undoing that gives an enlargement with scale factor $\frac{1}{2}$, centre $(-1, 7)$, which maps $U$ back onto $T$.

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

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

Write the coordinate rule for the first transformation, the rotation of 90 degrees clockwise about (0, 0).

For a rotation of $90^{\circ}$ clockwise about the point $(0, 0)$ the rule is $(x,\, y) \rightarrow (y,\, -x)$.

Apply the first transformation to vertex A.

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

Apply the first transformation to vertex B.

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

Apply the first transformation to vertex C.

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

State the intermediate image after the first transformation.

The intermediate triangle has vertices $A'(2,\, -1),\; B'(2,\, -3),\; C'(5,\, -3)$. The second transformation is applied to this triangle.

Write the coordinate rule for the second transformation, the translation by the vector (2, 4).

For a translation by the vector $\begin{pmatrix} 2 \\ 4 \end{pmatrix}$ the rule is $(x,\, y) \rightarrow (x + 2,\, y + 4)$.

Apply the second transformation to the image of A.

$(2, -1)$ maps to $(4, 3)$.

Apply the second transformation to the image of B.

$(2, -3)$ maps to $(4, 1)$.

Apply the second transformation to the image of C.

$(5, -3)$ maps to $(7, 1)$.

State the final image, triangle U.

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

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.

Reverse that single transformation.

The forward transformation is a rotation of $90^{\circ}$ clockwise about the point $(3, 1)$. Undoing it gives a rotation of $90^{\circ}$ anticlockwise about the point $(3, 1)$. The centre is the only point that does not move: $(3, 1)$.

Check the reversed transformation maps U back onto T.

Rotation of $90^{\circ}$ anticlockwise about the point $(3, 1)$ sends every vertex of $U$ back to the matching vertex of $T$.

State the answer.

Going forwards, $T$ maps onto $U$ by a rotation of $90^{\circ}$ clockwise about the point $(3, 1)$. Undoing that gives a rotation of $90^{\circ}$ anticlockwise about the point $(3, 1)$, which maps $U$ back onto $T$.