Combined Transformations and Invariance Revision Notes | GCSE AQA Higher Maths

Worked example: Combining a Reflection and a Translation

Triangle $P$ has vertices at $(1, 1)$, $(3, 1)$, and $(3, 4)$. Reflect triangle $P$ in the $y$-axis to give triangle $P'$. Then translate $P'$ by $\begin{pmatrix} -1 \\ -3 \end{pmatrix}$ to give triangle $P''$. Describe the changes and any invariant properties.

1. $(1, 1)$, $(3, 1)$, $(3, 4)$ — Always label vertices clearly before starting. [M1 — correct identification of object]
2. Rule: $(x, y) \rightarrow (-x, y)$ $(1, 1) \rightarrow (-1, 1)$ $(3, 1) \rightarrow (-3, 1)$ $(3, 4) \rightarrow (-3, 4)$ — Reflecting in the $y$-axis negates the $x$-coordinate and leaves $y$ unchanged. Triangle $P'$ has vertices $(-1, 1)$, $(-3, 1)$, $(-3, 4)$. [M1 — correct reflection rule applied]
3. Translation $\begin{pmatrix} -1 \\ -3 \end{pmatrix}$: subtract 1 from each $x$, subtract 3 from each $y$ $(-1, 1) \rightarrow (-2, -2)$ $(-3, 1) \rightarrow (-4, -2)$ $(-3, 4) \rightarrow (-4, 1)$ — Add the column vector components to each coordinate. $P''$ has vertices $(-2, -2)$, $(-4, -2)$, $(-4, 1)$. [M1 — correct translation applied to image, not object]
4. Changed: position (shape has moved), orientation (laterally inverted by the reflection) Invariant: side lengths, interior angles, area (shape is congruent to original) — The reflection changed orientation; the translation changed position further. Neither transformation changed size or angles. [A1 — full description of changes and invariance]

Answer: $P''$ has vertices $(-2, -2)$, $(-4, -2)$, $(-4, 1)$. The shape is congruent to the original. Orientation is reversed (laterally inverted). Position has changed. Side lengths and angles are invariant.

Worked example: Two Reflections in Parallel Lines Produce a Translation

Square $Q$ has vertices at $(1, 0)$, $(3, 0)$, $(3, 2)$, $(1, 2)$. Reflect $Q$ in the line $x = 4$ to give $Q'$. Then reflect $Q'$ in the line $x = 7$ to give $Q''$. Show that the combination is equivalent to a single translation and state the column vector.

1. Rule for reflection in $x = k$: $(x, y) \rightarrow (2k - x, y)$, here $k = 4$ $(1, 0) \rightarrow (2 \times 4 - 1, 0) = (7, 0)$ $(3, 0) \rightarrow (2 \times 4 - 3, 0) = (5, 0)$ $(3, 2) \rightarrow (5, 2)$ $(1, 2) \rightarrow (7, 2)$ — $Q'$ has vertices $(7, 0)$, $(5, 0)$, $(5, 2)$, $(7, 2)$. The shape is laterally inverted but congruent. [M1]
2. Now $k = 7$ $(7, 0) \rightarrow (2 \times 7 - 7, 0) = (7, 0)$ $(5, 0) \rightarrow (2 \times 7 - 5, 0) = (9, 0)$ $(5, 2) \rightarrow (9, 2)$ $(7, 2) \rightarrow (7, 2)$ — $Q''$ has vertices $(7, 0)$, $(9, 0)$, $(9, 2)$, $(7, 2)$. [M1]
3. Original $Q$: $(1,0),(3,0),(3,2),(1,2)$ $Q''$: $(7,0),(9,0),(9,2),(7,2)$ Change in $x$: $7 - 1 = 6$, $9 - 3 = 6$ Change in $y$: $0 - 0 = 0$ — Every vertex has moved exactly 6 units to the right and 0 units vertically. This is a pure translation. [M1]
4. Translation $\begin{pmatrix} 6 \\ 0 \end{pmatrix}$ Note: the distance between the parallel lines is $7 - 4 = 3$, and the translation distance is $2 \times 3 = 6$. — Two reflections in parallel lines always produce a translation. The translation distance equals twice the perpendicular distance between the mirror lines. [A1]

Answer: The combination of the two reflections is equivalent to a single translation of $\begin{pmatrix} 6 \\ 0 \end{pmatrix}$.

Worked example: Finding the Single Equivalent Transformation

Triangle $R$ has vertices $A(0, 2)$, $B(2, 2)$, $C(2, 5)$. It is rotated $90^\circ$ anticlockwise about the origin to give $R'$, then translated by $\begin{pmatrix} 3 \\ 1 \end{pmatrix}$ to give $R''$. Describe the single transformation that maps $R$ directly onto $R''$.

1. Rule for $90^\circ$ anticlockwise about origin: $(x, y) \rightarrow (-y, x)$ $A(0, 2) \rightarrow (-2, 0)$ $B(2, 2) \rightarrow (-2, 2)$ $C(2, 5) \rightarrow (-5, 2)$ — $R'$ has vertices $(-2, 0)$, $(-2, 2)$, $(-5, 2)$. [M1 — correct rotation rule]
2. Add $\begin{pmatrix} 3 \\ 1 \end{pmatrix}$ to each vertex: $(-2, 0) \rightarrow (1, 1)$ $(-2, 2) \rightarrow (1, 3)$ $(-5, 2) \rightarrow (-2, 3)$ — $R''$ has vertices $(1, 1)$, $(1, 3)$, $(-2, 3)$. [M1 — correct translation applied to $R'$]
3. Object $R$: $A(0,2)$, $B(2,2)$, $C(2,5)$ Image $R''$: $A''(1,1)$, $B''(1,3)$, $C''(-2,3)$ Orientation: check if $R''$ is a mirror image of $R$ — it is not, so the equivalent is a rotation (not a reflection). Look for a fixed point: test whether a rotation maps $R$ to $R''$ directly. — Since a rotation followed by a translation is generally equivalent to a single rotation (about a different centre), we need to find that centre. [M1 — correct reasoning about transformation type]
4. The rotation component is $90^\circ$ anticlockwise (the translation does not change the angle). To find the new centre: the perpendicular bisector of $AA''$ (from $(0,2)$ to $(1,1)$) and the perpendicular bisector of $BB''$ (from $(2,2)$ to $(1,3)$) must intersect at the centre. Midpoint of $AA''$: $\left(\frac{1}{2}, \frac{3}{2}\right)$. Gradient of $AA''$: $\frac{1-2}{1-0} = -1$. Perpendicular gradient: $1$. Line through midpoint: $y - \frac{3}{2} = 1\left(x - \frac{1}{2}\right) \Rightarrow y = x + 1$. Midpoint of $BB''$: $\left(\frac{3}{2}, \frac{5}{2}\right)$. Gradient of $BB''$: $\frac{3-2}{1-2} = -1$. Perpendicular gradient: $1$. Line through midpoint: $y - \frac{5}{2} = 1\left(x - \frac{3}{2}\right) \Rightarrow y = x + 1$. Both perpendicular bisectors are the same line — any point on $y = x + 1$ is equidistant. Test centre $(0, 1)$: rotate $A(0,2)$ by $90^\circ$ anticlockwise about $(0,1)$: translate to origin $(0,1)$, apply rotation, translate back. $(0,2) - (0,1) = (0,1) \rightarrow (-1, 0) \rightarrow (-1+0, 0+1) = (-1, 1)$. Not $(1,1)$. Test centre $(1, 2)$: $(0,2)-(1,2)=(-1,0) \rightarrow (0,-1) \rightarrow (0+1,-1+2)=(1,1)$. Correct! Verify with $B$: $(2,2)-(1,2)=(1,0) \rightarrow (0,1) \rightarrow (1,3)$. Correct. — The centre of the equivalent single rotation is $(1, 2)$. [A1 — correct centre identified with verification]
5. Single transformation: rotation of $90^\circ$ anticlockwise about the point $(1, 2)$. — This single rotation maps every point of $R$ directly onto $R''$, confirming it is the equivalent transformation. [A1]

Answer: A rotation of $90^\circ$ anticlockwise about the point $(1, 2)$.