When one transformation is followed by another, the result can very often be achieved by a single transformation instead. Finding that single transformation is the standard exam question.
- Apply the first transformation to get the intermediate shape. Label it clearly.
- Apply the second transformation to get the final image.
- Now ignore the middle shape completely and compare only the original with the final image.
- Check the size: same size means translation, reflection or rotation. Different size means enlargement.
- Check the orientation: has the shape been flipped over, or just turned?
- Describe the single transformation fully.
โข Not turned over โ it is a translation, a rotation, or an enlargement.
โข Turned over (mirror image) โ it involves a reflection.
Two reflections in intersecting mirrors $=$ a rotation about the crossing point
Intersecting mirrors. If the mirrors meet at an angle $\theta$, the rotation is through $2\theta$ about the point where they cross.
A shape is reflected in the line $x = 1$, and the image is then reflected in the line $x = 5$. Describe the single transformation equivalent to this pair.
A translation by $\begin{pmatrix} 8 \\ 0 \end{pmatrix}$.
Check with a point: $(3, 2)$ reflects in $x=1$ to $(-1, 2)$, which reflects in $x=5$ to $(11, 2)$. That is $8$ to the right โ
A shape is reflected in the $x$-axis and then in the $y$-axis. Describe the single equivalent transformation.
A rotation of $180^\circ$ about the origin $(0,\ 0)$.
This matches the rule: the mirrors meet at $90^\circ$, and $2 \times 90 = 180^\circ$ โ
| First | Then | Single equivalent |
|---|---|---|
| Translation | Translation | A translation โ add the two vectors |
| Rotation about $P$ | Rotation about the same $P$ | A rotation about $P$ โ add the angles |
| Reflection | Reflection (parallel mirrors) | A translation |
| Reflection | Reflection (intersecting mirrors) | A rotation about the crossing point |
| Enlargement SF $a$ | Enlargement SF $b$, same centre | An enlargement, SF $a \times b$ |
| Rotation | Translation | Usually a rotation through the same angle, about a different centre |
| Reflection | Translation parallel to the mirror | A "glide reflection" โ not a single named GCSE transformation |
A shape is translated by $\begin{pmatrix} 4 \\ -3 \end{pmatrix}$ and then by $\begin{pmatrix} -1 \\ 7 \end{pmatrix}$. Find the single equivalent translation.
A translation by $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$.
A shape is rotated $70^\circ$ clockwise about the origin, then $160^\circ$ clockwise about the origin. Describe the single equivalent transformation.
A rotation of $130^\circ$ anticlockwise about the origin (equivalently $230^\circ$ clockwise).
A shape is reflected in the line $y = x$ and then in the $x$-axis. Describe the single equivalent transformation.
A rotation of $90^\circ$ clockwise about $(0,\ 0)$.
Check with the mirror rule: $y = x$ and the $x$-axis meet at $45^\circ$, and $2 \times 45 = 90^\circ$ โ
Something is invariant under a transformation if it does not change.
| Property | Translation | Reflection | Rotation | Enlargement |
|---|---|---|---|---|
| Side lengths | Same | Same | Same | $\times k$ |
| Angles | Same | Same | Same | Same |
| Area | Same | Same | Same | $\times k^2$ |
| Orientation (way round) | Same | Reversed | Same | Same (reversed if $k \lt 0$) |
| Parallel lines stay parallel | Yes | Yes | Yes | Yes |
| Congruent to the original? | Yes | Yes | Yes | No |
An invariant point is a point that does not move โ its image is exactly where it started.
| Transformation | Invariant points |
|---|---|
| Translation (non-zero vector) | None โ everything moves |
| Reflection | Every point on the mirror line |
| Rotation | Only the centre of rotation |
| Enlargement ($k \neq 1$) | Only the centre of enlargement |
A rectangle has vertices $(1,\ 2)$, $(5,\ 2)$, $(5,\ 4)$ and $(1,\ 4)$. It is reflected in the line $y = 2$. How many of its vertices are invariant?
Two invariant vertices.
(In fact the whole edge joining them is invariant โ every point on it stays put.)
Triangle $T$ has vertices $(2,\ 1)$, $(6,\ 1)$ and $(2,\ 5)$. Find a reflection under which exactly one vertex of $T$ is invariant.
A reflection in $x = 6$ leaves exactly one vertex invariant.
Another valid answer: $y = 5$, which passes only through $(2,\ 5)$.
Method
Do both, then ignore the middle shape and compare start with finish.
Order matters
$A$ then $B$ is generally not the same as $B$ then $A$.
Parallel mirrors
Give a translation of $2d$, where $d$ is the gap.
Intersecting mirrors
Give a rotation of $2\theta$ about the crossing point.
Two translations
Add the vectors.
Two rotations
Same centre: add the angles.
Algebraic check
Track $(x,y)$ through both steps; the final rule names the transformation.
Always invariant
Angles, under every one of the four transformations.
Invariant points
Mirror line; centre of rotation; centre of enlargement; none for a translation.
A shape is translated by $\begin{pmatrix} -2 \\ 5 \end{pmatrix}$ and then by $\begin{pmatrix} 7 \\ -9 \end{pmatrix}$. Find the single equivalent translation.
โถ Show solution
Add the components: $-2 + 7 = 5$ and $5 + (-9) = -4$.
A translation by $\begin{pmatrix} 5 \\ -4 \end{pmatrix}$.
A shape is reflected in the $y$-axis and then in the $x$-axis. Describe the single equivalent transformation.
โถ Show solution
$(x,\ y) \to (-x,\ y) \to (-x,\ -y)$
The rule $(x,y) \to (-x,-y)$ is a rotation of $180^\circ$ about the origin.
(The mirrors meet at $90^\circ$, and $2 \times 90 = 180^\circ$ โ)
A shape is reflected in $y = 2$ and then in $y = 6$. Describe the single equivalent transformation.
โถ Show solution
The mirrors are parallel (both horizontal), $6 - 2 = 4$ apart.
Translation distance $= 2 \times 4 = 8$, in the direction from the first mirror to the second, i.e. upwards.
A translation by $\begin{pmatrix} 0 \\ 8 \end{pmatrix}$.
How many invariant points are there when a triangle with vertices $(0,\ 0)$, $(4,\ 0)$ and $(0,\ 3)$ is reflected in the $x$-axis?
โถ Show solution
Invariant points under a reflection are those on the mirror line, here $y = 0$.
$(0,\ 0)$ has $y = 0$ โ; $(4,\ 0)$ has $y = 0$ โ; $(0,\ 3)$ does not.
Two invariant vertices โ and in fact the whole side joining them is invariant.
A shape is enlarged by scale factor $3$ about the origin, then by scale factor $\tfrac{1}{2}$ about the origin. Describe the single equivalent transformation.
โถ Show solution
Multiply the scale factors: $3 \times \tfrac{1}{2} = 1.5$.
An enlargement of scale factor $1.5$, centre the origin.
A shape is rotated $90^\circ$ clockwise about the origin and then reflected in the $x$-axis. Find the single equivalent transformation.
โถ Show solution
$90^\circ$ clockwise: $(x,\ y) \to (y,\ -x)$
Reflect in the $x$-axis: $(y,\ -x) \to (y,\ x)$
The rule $(x,y) \to (y,x)$ is a reflection in the line $y = x$.
Under which transformations is the area of a shape invariant? Explain your answer.
โถ Show solution
Area is invariant under translation, reflection and rotation, because these all produce a congruent image โ the shape is moved but never resized.
Area is not invariant under an enlargement (unless the scale factor is $1$ or $-1$): lengths are multiplied by $k$, so area is multiplied by $k^2$.
Square $S$ has vertices $(1,\ 1)$, $(3,\ 1)$, $(3,\ 3)$ and $(1,\ 3)$. It is rotated $90^\circ$ anticlockwise about $(2,\ 2)$. Describe what happens to the square, and state the invariant points.
โถ Show solution
$(2,\ 2)$ is the centre of the square, so rotating $90^\circ$ about it maps the square exactly onto itself โ each vertex moves round to the next one.
The shape as a whole is invariant, but the only invariant point is the centre $(2,\ 2)$: every other point has moved.
A shape is reflected in the line $y = x$ and then in the $y$-axis. Use algebra to find the single equivalent transformation.
โถ Show solution
Reflect in $y = x$: $(x,\ y) \to (y,\ x)$
Reflect in the $y$-axis (change the sign of the first coordinate): $(y,\ x) \to (-y,\ x)$
The rule $(x,y) \to (-y,\ x)$ is a rotation of $90^\circ$ anticlockwise about the origin.
Check with the mirror rule: $y = x$ and the $y$-axis meet at $45^\circ$, and $2 \times 45 = 90^\circ$ โ
Triangle $A$ has vertices $(1,\ 1)$, $(3,\ 1)$ and $(1,\ 2)$.
(a) Reflect $A$ in the $y$-axis to give $B$; write down $B$'s vertices.
(b) Translate $B$ by $\begin{pmatrix} 0 \\ -4 \end{pmatrix}$ to give $C$; write down $C$'s vertices.
(c) Describe fully the single transformation mapping $A$ to $C$, or explain why no single named transformation does it.
โถ Show solution
(a) Reflecting in the $y$-axis changes the sign of $x$:
$B$: $(-1,\ 1)$, $(-3,\ 1)$, $(-1,\ 2)$
(b) Subtract $4$ from each $y$:
$C$: $(-1,\ -3)$, $(-3,\ -3)$, $(-1,\ -2)$
(c) Compare $A$ and $C$. The triangle has been turned over (a reflection is involved) and also shifted, so it is not a pure reflection, rotation or translation.
Test a reflection: a single mirror would need to be the perpendicular bisector of every join. $(1,1) \to (-1,-3)$ has midpoint $(0,-1)$; $(3,1)\to(-3,-3)$ has midpoint $(0,-1)$; but $(1,2)\to(-1,-2)$ has midpoint $(0,0)$. The midpoints differ, so no single mirror works.
Test a rotation of $180^\circ$ about $(0,-1)$: that maps $(1,1)\to(-1,-3)$ โ and $(3,1)\to(-3,-3)$ โ but $(1,2)\to(-1,-4)$ โ.
Conclusion: the combination is a glide reflection โ a reflection followed by a translation parallel to the mirror. This is not one of the four named GCSE transformations, so no single one describes it.