G7, G24 - Reflections, rotations, translations and enlargements

A shape can move, turn, reflect or change size in a precisely controlled way. You will construct images vertex by vertex, describe each transformation completely, use column vectors for translations, and recognise which properties stay unchanged.

Reflections in a mirror line

Can an enlargement make a shape smaller? Keep that question in mind as you first explore transformations that preserve its size.

You already know

A coordinate pair $(x,y)$ gives horizontal position first, then vertical position. Congruent shapes have the same size and shape, even if one faces a different way.

A reflection puts a point and its image at equal perpendicular distances on opposite sides of a mirror line. The mirror is the perpendicular bisector of each joining segment $PP'$ . Points on the mirror stay fixed.

To construct a polygon's reflection, draw a perpendicular from each vertex to the mirror. Count or measure its distance, continue by the same distance on the other side, and mark the image with a prime. Join image vertices in the original order. With a sloping mirror, a horizontal move usually fails because the distance must be perpendicular.

Reflection in the $x$ -axis sends $(x,y)$ to $(x,-y)$ ; in the $y$ -axis it gives $(-x,y)$ . Reflection in $y=x$ swaps coordinates, and in $y=-x$ it gives $(-y,-x)$ . These rules are not on the formula sheet: learn them, and check using equal perpendicular distances.

[DIAGRAM: asset_name: Reflections in vertical and diagonal mirror lines; asset_slug: 1ma1_m49_reflections; caption: Each reflected vertex lies at the same perpendicular distance on the opposite side of its mirror.; description: Two coordinate panels show triangle A(3,1), B(5,1), C(3,3) reflected in x equals 2, and triangle P(1,3), Q(2,5), R(3,3) reflected in y equals x. Dashed connecting segments are perpendicular to the mirror lines, and images carry prime labels.]
Each reflected vertex lies at the same perpendicular distance on the opposite side of its mirror.

The diagonal panel deliberately includes a vertex on the mirror: $R=(3,3)$ stays fixed. Every other vertex still needs its own reflected position.

Worked example 1: a vertical mirror

Question. Reflect $P=(5,1)$ in $x=2$ .

Plan. Find the horizontal distance to this vertical mirror, then go equally far beyond it.

\begin{aligned} \text{distance to mirror}&=5-2=3 &&\text{horizontal perpendicular}\\ x'&=2-3=-1 &&\text{same distance on other side}\\ y'&=1 &&\text{unchanged height} \end{aligned}

Why subtract from 2? The image lies 3 units to the left of the mirror, not 3 units left of the original point.

Answer. $P'=(-1,1)$ .

Check. The midpoint has $x=(5+(-1))/2=2$ , on the mirror.

Change only the mirror to $x=-1$ . The original point stays fixed, but the image must be equally far from the new line.

Worked example 2: moving the mirror

Question. Reflect $P=(5,1)$ in $x=-1$ .

Plan. Measure from the new mirror, then cross it by the same distance.

\begin{aligned} \text{distance to mirror}&=5-(-1)=6 &&\text{perpendicular distance}\\ x'&=-1-6=-7 &&\text{opposite side}\\ y'&=1 &&\text{vertical mirror} \end{aligned}

Why does the image move farther left? The mirror has moved left, increasing the original point's distance from it.

Answer. $P'=(-7,1)$ .

Check. $(5+(-7))/2=-1$ , the new mirror position.

For a diagonal example, $Q=(-1,4)$ reflected in $y=x$ becomes $(4,-1)$ . Their midpoint $(1.5,1.5)$ lies on the mirror, and the joining segment is perpendicular to it.

To identify an unknown reflection, construct the perpendicular bisector of a corresponding-vertex joining segment, then check a second pair. Give the mirror's equation, such as “reflection in $x=2$ ”. On squared paper, practise reflecting triangle $(1,1),(4,1),(2,3)$ separately in $x=-1$ and $y=x$ , using perpendicular distances. A typed coordinate answer does not check drawing accuracy.

Watch out

Wrong: reflect $(5,1)$ in $x=2$ by moving 3 units left to $(2,1)$ . That stops at the mirror. Continue another 3 units to $(-1,1)$ , making the mirror the midpoint.

Use a new mirror position for your turn.

Rotations about a centre

A rotation turns every point through the same angle and direction about a fixed centre.

A complete description gives centre, angle and clockwise or anticlockwise direction. A $180^\circ$ turn gives the same image in either direction, so its direction is unnecessary. The centre stays fixed; other points travel round circles about it. Rotations preserve lengths, angles and the order of labelled vertices.

To construct $P'$ , join $OP$ , turn a ray through the required angle at $O$ , then mark $OP'=OP$ on the turned ray. Use a protractor for the angle and a compass for the radius. Repeat for each vertex. On a square grid, count offsets for quarter-turns and half-turns.

About the origin, $90^\circ$ anticlockwise sends $(x,y)$ to $(-y,x)$ ; clockwise gives $(y,-x)$ ; $180^\circ$ gives $(-x,-y)$ . These rules are not on the formula sheet: learn them. For another centre, subtract its coordinates before turning the offset and add them back afterwards.

Worked example 3: an anticlockwise quarter-turn

Question. Rotate $P=(4,3)$ through $90^\circ$ anticlockwise about $O=(1,1)$ .

Plan. Find the offset from $O$ , turn it, then return to the grid's coordinates.

\begin{aligned} P-O&=(3,2) &&\text{subtract the centre}\\ (3,2)&\mapsto(-2,3) &&\text{anticlockwise quarter-turn}\\ P'&=(1-2,1+3)=(-1,4) &&\text{add the centre back} \end{aligned}

Why turn the offset? The rotation is about $O$ , rather than the origin.

Answer. $P'=(-1,4)$ .

Check. Both squared radii equal $3^2+2^2=13$ , so the distance from the centre is unchanged.

Change only the direction to clockwise. The radius stays the same, but the point turns to the other side.

Worked example 4: reversing the direction

Question. Rotate $P=(4,3)$ through $90^\circ$ clockwise about $O=(1,1)$ .

Plan. Turn the same offset clockwise instead.

\begin{aligned} P-O&=(3,2) &&\text{centre-relative offset}\\ (3,2)&\mapsto(2,-3) &&\text{clockwise quarter-turn}\\ P'&=(1+2,1-3)=(3,-2) &&\text{return to coordinates} \end{aligned}

Why does the second component become negative? An offset pointing right turns down in a clockwise quarter-turn.

Answer. $P'=(3,-2)$ .

Check. The squared radius remains $2^2+(-3)^2=13$ ; turning the offset back recovers $(3,2)$ .

For a half-turn, $Q=(-3,4)$ about $C=(2,-1)$ has offset $(-5,5)$ , which becomes $(5,-5)$ . Adding $C$ gives $Q'=(7,-6)$ . Their midpoint is $C$ , checking the centre.

To find an unknown rotation centre, intersect perpendicular bisectors of two corresponding-vertex joining segments. Check the angle and direction from that centre. For a half-turn, every pair's midpoint is the centre. On paper, rotate triangle $(2,1),(4,1),(2,3)$ by $90^\circ$ anticlockwise about $(1,1)$ . The image is $(1,2),(1,4),(-1,2)$ , as shown in the later diagram.

Watch out

Wrong: rotate $(4,3)$ about $(1,1)$ by directly writing $(-3,4)$ . That rule turns about the origin. Subtract $(1,1)$ , turn the offset, then add $(1,1)$ back.

Use the clockwise method again with new coordinates.

Translations and column vectors

A translation moves every point by the same horizontal and vertical amounts, without turning or changing the shape.

Column vector

The translation vector

\begin{pmatrix}a\\b\end{pmatrix}

means move $a$ horizontally and $b$ vertically. Positive entries mean right and up; negative entries mean left and down. A vector is a displacement, not a point's position.

Thus $\begin{pmatrix}-3\\2\end{pmatrix}$ means 3 left and 2 up wherever a vertex starts. Add the top entry to every $x$ -coordinate and the bottom entry to every $y$ -coordinate. Join the image vertices in order. This coordinate rule is not on the formula sheet: learn it.

Worked example 5: translating a triangle

Question. Translate $A=(1,-2)$ , $B=(4,-2)$ , $C=(2,1)$ by $\begin{pmatrix}-3\\2\end{pmatrix}$ .

Plan. Add the same two components to each vertex.

\begin{aligned} A'&=(1-3,-2+2)=(-2,0) &&\text{left }3\text{ and up }2\\ B'&=(4-3,-2+2)=(1,0) &&\text{same displacement}\\ C'&=(2-3,1+2)=(-1,3) &&\text{same displacement} \end{aligned}

Why use the same move three times? Different moves would change the relative positions of the vertices and distort the triangle.

Answer. $A'=(-2,0)$ , $B'=(1,0)$ , $C'=(-1,3)$ .

Check. Each image minus original gives $(-3,2)$ .

Last Accessed

Papers 1, 2 and 3

G7, G24 - Reflections, rotations, translations and enlargements

Reflections in a mirror line

Rotations about a centre

Translations and column vectors

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