HG7–HG8 - Negative enlargements and combined transformations | Eduqas GCSE Mathematics

Learn how a negative scale factor places an image beyond the centre, and how the order of transformations affects the result. HG7 develops negative enlargement; HG8 concerns combined movements, their invariants and equivalent single transformations.

Negative scale factors

Does scale factor $-2$ mean an image has negative side lengths? No: the minus sign reverses the displacement from the centre, while the magnitude $2$ doubles every length.

Negative enlargement

For centre $O$ and scale factor $k<0$ , multiply each displacement from $O$ by $k$ . The image lies on the opposite side of $O$ along the same straight line. Distances from the centre and side lengths multiply by $|k|$ ; corresponding angles remain equal.

You can view a negative enlargement as a positive enlargement of factor $|k|$ followed by a half-turn about the centre. It is not a mirror reflection: in the plane a half-turn preserves the clockwise order of the vertices. Factor $-1$ is exactly a $180^\circ$ rotation about the centre and produces a congruent image.

Worked example 1: a negative integer factor

Question. Enlarge $P(3,2)$ , $Q(5,2)$ , $R(3,4)$ by factor $-2$ , centre $O(1,1)$ .

Plan. Multiply each displacement from $O$ by $-2$ , then add the centre back.

Vertex	Displacement from $O$	Multiply by $-2$	Image
$P(3,2)$	$(2,1)$	$(-4,-2)$	$P'(-3,-1)$
$Q(5,2)$	$(4,1)$	$(-8,-2)$	$Q'(-7,-1)$
$R(3,4)$	$(2,3)$	$(-4,-6)$	$R'(-3,-5)$

Answer. $P'(-3,-1),Q'(-7,-1),R'(-3,-5)$ .

Check. The image side $P'Q'$ has length $4$ , twice the original $2$ . Along every point-to-image line, $O$ lies between the points and the image is twice as far from $O$ .

Keep the original and centre, but change the magnitude of the negative factor to $1/2$ . The direction still reverses; the image becomes smaller.

Worked example 2: a negative fractional factor

Question. Enlarge the same triangle by factor $-1/2$ , centre $O(1,1)$ .

Plan. Halve each centre-relative displacement and reverse its direction.

\begin{aligned} (2,1)&\longmapsto(-1,-0.5)\quad\Longrightarrow\quad P'(0,0.5)\\ (4,1)&\longmapsto(-2,-0.5)\quad\Longrightarrow\quad Q'(-1,0.5)\\ (2,3)&\longmapsto(-1,-1.5)\quad\Longrightarrow\quad R'(0,-0.5) \end{aligned}

Answer. $P'(0,0.5),Q'(-1,0.5),R'(0,-0.5)$ .

Check. Reversing and doubling each image displacement from $O$ recovers the original. The image side $P'Q'$ has length $1$ , half the original $2$ .

The centre separates each original vertex from its image in both cases.

[DIAGRAM: asset_name: Negative scale factors through the centre; asset_slug: negative-factor-pair; caption: The sign changes the side of the centre; the magnitude controls size.; description: Exact coordinate panels enlarge P(3,2), Q(5,2), R(3,4) about O(1,1) by factors minus 2 and minus one half. Original and image vertices are joined through O. Correct images are (-3,-1),(-7,-1),(-3,-5) and (0,0.5),(-1,0.5),(0,-0.5).]
The sign changes the side of the centre

Watch out

Do not multiply absolute coordinates unless the centre is the origin. For centre $(1,1)$ , factor $-2$ sends $(3,2)$ to $(-3,-1)$ , not $(-6,-4)$ .

Use a fresh triangle and centre.

Combining transformations

A sequence of transformations acts on the current image, in the stated order. The second movement uses the first image as its input.

Keep intermediate labels such as $P'$ , then use $P''$ for the final image. This makes it possible to check each stage rather than guess a direct movement.

Worked example 3: reflect, then translate

Question. Triangle $P(2,1),Q(3,1),R(2,3)$ is reflected in the $y$ -axis, then translated by $\begin{pmatrix}2\\0\end{pmatrix}$ . Find the final image.

Plan. Reflect every vertex first, then add the translation to that reflected image.

Vertex	After reflection	After translation
$P(2,1)$	$P'(-2,1)$	$P''(0,1)$
$Q(3,1)$	$Q'(-3,1)$	$Q''(-1,1)$
$R(2,3)$	$R'(-2,3)$	$R''(0,3)$

Answer. $P''(0,1),Q''(-1,1),R''(0,3)$ .

Check. Undo the translation, then reflect again. For example, $(0,1)\mapsto(-2,1)\mapsto(2,1)$ returns $P$ .

Now change only the order. A reflection also reverses the direction of a horizontal displacement made before it.

Worked example 4: translate, then reflect

Question. Translate the original triangle by $\begin{pmatrix}2\\0\end{pmatrix}$ , then reflect it in the $y$ -axis.

Plan. Add the vector first, then reflect the translated coordinates.

Vertex	After translation	After reflection
$P(2,1)$	$P'(4,1)$	$P''(-4,1)$
$Q(3,1)$	$Q'(5,1)$	$Q''(-5,1)$
$R(2,3)$	$R'(4,3)$	$R''(-4,3)$

Answer. $P''(-4,1),Q''(-5,1),R''(-4,3)$ , different from the first order.

Check. Reflecting the final image and translating by $\begin{pmatrix}-2\\0\end{pmatrix}$ returns the original. Both final triangles remain congruent to it.

Rotations follow the same stage-by-stage rule, retaining the stated rotation centre.

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HG7–HG8 - Negative enlargements and combined transformations

Negative scale factors

Combining transformations

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