5.2 - Rotations, reflections and translations

Move shapes by translation, reflection and rotation, construct their images on paper and give complete descriptions of the movements. You will use corresponding vertices to check both the position and the unchanged shape and size.

Translations and column vectors

Is “turn the triangle through 90 degrees” enough information to locate its image? There are many places a triangle could go. We will learn which information fixes each movement, starting with a slide: 4 right and 2 down.

You already know

A coordinate pair gives horizontal position first, then vertical position. Congruent shapes have the same shape and size. Corresponding vertices are matching points, such as $A$ and its image $A'$ .

A slide moves every vertex in the same way. It does not turn the shape or change which way it faces.

Translation

A translation moves every point through the same distance in the same direction. The column vector $\begin{pmatrix}h\\v\end{pmatrix}$ records the horizontal movement first and the vertical movement second. Right and up are positive; left and down are negative.

Thus translation by $\begin{pmatrix}h\\v\end{pmatrix}$ sends $(x,y)$ to $(x+h,y+v)$ . The two components specify one movement, not two transformations. This rule is not in either the Foundation or Higher formulae sheet: learn it.

Worked example 1: translating a triangle

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

Plan. Add 4 to each horizontal coordinate and subtract 2 from each vertical coordinate.

\begin{aligned} A'&=(-2+4,1-2)=(2,-1) &&\text{move }A\\ B'&=(1+4,1-2)=(5,-1) &&\text{use the same movement}\\ C'&=(-2+4,3-2)=(2,1) &&\text{move the final vertex} \end{aligned}

Why the same changes each time? A translation slides the whole shape without turning or distorting it.

Answer. $A'(2,-1)$ , $B'(5,-1)$ , $C'(2,1)$ .

Check. Each image-minus-object difference is $(4,-2)$ ; side $AB$ and side $A'B'$ both have length 3.

Change only the sign of the vertical component. The image now moves up instead of down.

Worked example 2: changing the vertical direction

Question. Translate the same triangle by $\begin{pmatrix}4\\2\end{pmatrix}$ .

Plan. Add 4 horizontally and 2 vertically to every vertex.

\begin{aligned} A'&=(-2+4,1+2)=(2,3) &&\text{right and up}\\ B'&=(1+4,1+2)=(5,3) &&\text{same movement}\\ C'&=(-2+4,3+2)=(2,5) &&\text{same movement} \end{aligned}

Why is the shape still facing the same way? Every point has received exactly the same displacement.

Answer. $A'(2,3)$ , $B'(5,3)$ , $C'(2,5)$ .

Check. Each pair differs by $(4,2)$ , and the horizontal and vertical side lengths remain 3 and 2.

To draw either image on squared paper, plot and label the three calculated vertices, then join them in the original order. To identify a translation from a drawing, subtract the coordinates of an object vertex from its matching image vertex and check a second pair.

Watch out

$\begin{pmatrix}4\\-2\end{pmatrix}$ does not mean 4 down and 2 left. The top entry is horizontal: move 4 right and 2 down.

Change just the horizontal sign of the second example's vector.

Reflections and mirror lines

A mirror image is placed by distance from a line, rather than distance from a point. Each object-image join crosses the mirror at a right angle and is cut in half by it.

Reflection

A reflection places a point the same perpendicular distance on the opposite side of a mirror line. Points on the mirror remain fixed. A complete description names the reflection and its mirror line, for example “reflection in $x=1$ ”.

For $x=c$ , horizontal coordinates average to $c$ , so $(x,y)$ becomes $(2c-x,y)$ . For $y=d$ , the image is $(x,2d-y)$ . Reflection in $y=x$ , also written $y-x=0$ , swaps the coordinates. Reflection in $y=-x$ gives $(-y,-x)$ . These rules are not in either tier's formulae sheet: understand them through equal perpendicular distances.

Worked example 3: a vertical mirror

Question. Reflect $A(-2,1)$ , $B(1,1)$ , $C(-2,3)$ in $x=1$ .

Plan. Keep each height and place the point equally far across the vertical mirror.

\begin{aligned} A'&=(2\times1-(-2),1)=(4,1) &&\text{3 units across from the mirror}\\ B'&=(2\times1-1,1)=(1,1) &&\text{on the mirror, so fixed}\\ C'&=(2\times1-(-2),3)=(4,3) &&\text{keep the height} \end{aligned}

Why is $B$ unchanged? Its perpendicular distance to the mirror is zero.

Answer. $A'(4,1)$ , $B'(1,1)$ , $C'(4,3)$ .

Check. The midpoints of $AA'$ and $CC'$ have horizontal coordinate 1, and both joins are horizontal, perpendicular to the mirror.

Change only the mirror line to $y=x$ . Keep the same object so you can see the effect of a diagonal mirror.

Worked example 4: a diagonal mirror

Question. Reflect the same triangle in $y=x$ .

Plan. Swap the two coordinates of every vertex.

\begin{aligned} A(-2,1)&\mapsto A'(1,-2) &&\text{swap coordinates}\\ B(1,1)&\mapsto B'(1,1) &&\text{point on the mirror}\\ C(-2,3)&\mapsto C'(3,-2) &&\text{swap coordinates} \end{aligned}

Why does swapping work? A point and its swapped image have a midpoint on $y=x$ , and their join is perpendicular to that line.

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

Check. The midpoint of $AA'$ is $(-\frac12,-\frac12)$ , on $y=x$ ; the join has gradient $-1$ , perpendicular to the mirror's gradient 1.

To draw a reflection on paper, draw a perpendicular from each vertex to the mirror, mark the same distance beyond it, and join the image vertices. Diagonal mirrors require perpendicular distances, not simply horizontal gaps.

To construct a mirror line from an object and image, use a noncoincident matching pair. Draw equal compass arcs from its two endpoints with radius greater than half their separation. Make the arcs intersect on both sides, then join the two intersections with a ruler. This constructs the perpendicular bisector. Retain the arcs and check a second pair. For $P(0,1)$ , $P'(4,1)$ , $Q(1,3)$ , $Q'(3,3)$ , the midpoints $(2,1)$ , $(2,3)$ and horizontal joins identify the mirror $x=2$ . A fixed point alone does not determine the mirror.

Watch out

Reflection in $y=x$ is not $(x,y)\mapsto(-x,-y)$ ; that is a half turn about the origin. Reflection in $y=x$ swaps coordinates.

Change the mirror once more. All information needed is in the question.

Rotations about a centre

A rotation turns a radius about a fixed point while keeping its length. Changing that fixed point changes the image, even when the angle stays the same.

Rotation

A rotation moves every point through the same angle and direction about a fixed centre. Positive angles are anticlockwise; negative angles are clockwise. Thus $+90^\circ$ is a quarter turn anticlockwise and $-90^\circ$ is a quarter turn clockwise. A $180^\circ$ turn gives the same image in either direction.

On paper, join the centre to a vertex. Use a protractor to turn that radius through the given angle in the correct direction. Mark the same radius length on the new ray. Repeat for every vertex and join the images. Tracing paper can check the turn. The centre itself stays fixed.

About the origin, an anticlockwise quarter turn sends $(x,y)$ to $(-y,x)$ ; a clockwise quarter turn gives $(y,-x)$ ; a half turn gives $(-x,-y)$ . For another centre, apply these rules to the movement from the centre, then add its coordinates back. None of these rules is supplied in either tier's formulae sheet.

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Number

Algebra

Shape, space and measure

Handling data

5.2 - Rotations, reflections and translations

Translations and column vectors

Reflections and mirror lines

Rotations about a centre

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