Negative scale factors and finding the centre – Enlargement and Similarity in 2D (Senior 3 Mathematics) | Ganzaa

Enlargement and Similarity in 2D

Lessons

Senior 3 · Mathematics

Enlargement and Similarity in 2D

5 lessons

Lesson 3 of 5

Negative scale factors and finding the centre

25 min

An upside-down picture

Make a small hole in one side of a closed box and look at the inside of the opposite side: a faint picture of the scene outside appears there, upside down. Light from the top of a tree travels in a straight line through the hole and lands low on the back of the box; light from the foot of the tree lands higher up. The hole acts as a centre of enlargement, and the picture is on the other side of the centre from the tree. That is an enlargement with a negative scale factor.

In the drawing, the image is half as tall as the tree and on the opposite side of the pinhole: the scale factor is $-\frac{105}{210} = -\frac{1}{2}$ .

Definition

An enlargement with a negative scale factor $k$ (centre $O$ ) still follows $\overrightarrow{OP'} = k\,\overrightarrow{OP}$ . Because $k < 0$ :

$P^{'}$ is on the line $O P$ but on the opposite side of $O$ from $P$ , with $OP' = |k| \times OP$ ;
the image is similar to the object with ratio $∣ k ∣$ , and is turned through $180^\circ$ (upside down).

Example: $k = - 1$ gives $\overrightarrow{OP'} = -\overrightarrow{OP}$ : this is the central symmetry (half-turn) you met in Senior 2, an enlargement that keeps the size.

Important

The coordinate rules do not change: centre the origin, $(x, y) \mapsto (kx, ky)$ ; centre $X (a, b)$ , $(x, y) \mapsto \big(a + k(x - a),\; b + k(y - b)\big)$ . Just keep the negative signs.

Worked example 1: scale factor $- 2$ from the origin

Enlarge triangle $A (1, 2)$ , $B (3, 2)$ , $C (3, 3)$ with centre $O$ and scale factor $- 2$ .

Solution Multiply each coordinate by $- 2$ : $A^{'} (- 2, - 4)$ , $B^{'} (- 6, - 4)$ , $C^{'} (- 6, - 6)$ . The image is twice as large, on the far side of $O$ , and upside down: $C$ is above $B$ , but $C^{'}$ is below $B^{'}$ .

Worked example 2: a fractional negative scale factor

Enlarge triangle $P (3, 5)$ , $Q (7, 5)$ , $R (7, 3)$ with centre $X (1, 1)$ and scale factor $-\frac{1}{2}$ .

Solution $\overrightarrow{XP} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}$ , so $\overrightarrow{XP'} = -\frac{1}{2}\begin{pmatrix} 2 \\ 4 \end{pmatrix} = \begin{pmatrix} -1 \\ -2 \end{pmatrix}$ and $P^{'} (0, - 1)$ . Likewise $\overrightarrow{XQ} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}$ gives $Q^{'} (- 2, - 1)$ , and $\overrightarrow{XR} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}$ gives $R^{'} (- 2, 0)$ .

Finding the centre and the scale factor

Given an object and its image, you can work backwards.

Centre: draw straight lines through at least two pairs of matching points ( $A$ and $A^{'}$ , $B$ and $B^{'}$ ). They meet at the centre.
Size of $k$ : divide an image length by the matching object length.
Sign of $k$ : positive if object and image are on the same side of the centre (same way up); negative if the centre lies between them (image upside down).

With coordinates: from $\overrightarrow{XA'} = k\,\overrightarrow{XA}$ , the centre is $X = \frac{A' - kA}{1 - k}$ , worked out for each coordinate ( $k \neq 1$ ).

Worked example 3: a positive scale factor

Triangle $D (3, 2)$ , $E (5, 2)$ , $F (3, 3)$ is mapped to $D^{'} (5, 4)$ , $E^{'} (11, 4)$ , $F^{'} (5, 7)$ . Find the centre and the scale factor.

Solution $\overrightarrow{D'E'} = \begin{pmatrix} 6 \\ 0 \end{pmatrix} = 3\,\overrightarrow{DE}$ , and the image is the same way up, so $k = 3$ .

Centre: $x = \frac{5 - 3 \times 3}{1 - 3} = \frac{-4}{-2} = 2$ and $y = \frac{4 - 3 \times 2}{1 - 3} = \frac{-2}{-2} = 1$ , so $X (2, 1)$ . The figure confirms it: the lines $D D^{'}$ , $E E^{'}$ and $F F^{'}$ all pass through $(2, 1)$ .

Worked example 4: a negative scale factor

Triangle $A (5, 3)$ , $B (7, 3)$ , $C (5, 4)$ is mapped to $A^{'} (2, 3)$ , $B^{'} (- 2, 3)$ , $C^{'} (2, 1)$ . Describe the enlargement fully.

Solution $\overrightarrow{AB} = \begin{pmatrix} 2 \\ 0 \end{pmatrix}$ and $\overrightarrow{A'B'} = \begin{pmatrix} -4 \\ 0 \end{pmatrix} = -2\,\overrightarrow{AB}$ , so $k = - 2$ (twice as long, pointing the opposite way).

Centre: $X = \frac{A' - (-2)A}{1 - (-2)} = \frac{(2, 3) + (10, 6)}{3} = (4, 3)$ . Check with the lines: $A A^{'}$ is the line $y = 3$ , and $C C^{'}$ is the line $y = x - 1$ ; they meet at $(4, 3)$ . ✓

Full description: an enlargement with centre $(4, 3)$ and scale factor $- 2$ .

Warning

Always give all three facts when asked to describe an enlargement: the word "enlargement", the centre, and the scale factor with its sign. "Scale factor 2" for Worked example 4 loses marks: it is $- 2$ .

A negative scale factor does not mean a smaller image. The size depends on $∣ k ∣$ ; the sign only says which side of the centre.

Tip

If the lines joining matching points cross between the object and the image, $k$ is negative. If they have to be extended beyond both shapes to meet, $k$ is positive.

Corbettmaths: Enlargements with negative scale factors. Watch how each vertex is taken through the centre and out the other side, as in Worked example 1.

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Practice

The last two are harder.

Find the images of $(1, 2)$ and $(2, - 1)$ under the enlargement with centre the origin and scale factor $- 3$ .
Enlarge the triangle $(1, 1)$ , $(4, 1)$ , $(1, 3)$ with centre the origin and scale factor $- 1$ . Which isometry is this?
Find the image of $(3, 4)$ under the enlargement with centre $(2, 2)$ and scale factor $- 2$ .
Find the image of $(6, - 4)$ under the enlargement with centre the origin and scale factor $-\frac{1}{2}$ .
$A (2, 1)$ and $B (3, 1)$ are mapped to $A^{'} (3, 3)$ and $B^{'} (5, 3)$ by an enlargement. Find its centre and scale factor.
$P (1, 2)$ and $Q (3, 2)$ are mapped to $P^{'} (7, - 1)$ and $Q^{'} (3, - 1)$ . Describe the enlargement fully.
A triangle is enlarged with centre $O$ and scale factor $- 3$ . A side of the object is 2.5 cm long. How long is the matching side of the image, and how is the image placed?
Show that the enlargement with centre $O$ and scale factor $- 1$ has the same effect as the rotation of $180^\circ$ about $O$ .

Show answersHide answers

$(- 3, - 6)$ and $(- 6, 3)$ .
$(- 1, - 1)$ , $(- 4, - 1)$ , $(- 1, - 3)$ : the half-turn (central symmetry) about $O$ .
$\overrightarrow{XP} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ , so $\overrightarrow{XP'} = \begin{pmatrix} -2 \\ -4 \end{pmatrix}$ and the image is $(0, - 2)$ .
$(- 3, 2)$ .
$\overrightarrow{A'B'} = \begin{pmatrix} 2 \\ 0 \end{pmatrix} = 2\,\overrightarrow{AB}$ , so $k = 2$ ; centre $\frac{(3, 3) - 2(2, 1)}{1 - 2} = \frac{(-1, 1)}{-1} = (1, -1)$ .
$\overrightarrow{P'Q'} = \begin{pmatrix} -4 \\ 0 \end{pmatrix} = -2\,\overrightarrow{PQ}$ , so $k = - 2$ ; centre $\frac{(7, -1) + 2(1, 2)}{3} = (3, 1)$ . An enlargement with centre $(3, 1)$ and scale factor $- 2$ .
$3 \times 2.5 = 7.5$ cm; the image is on the other side of $O$ and upside down.
With $k = - 1$ , $(x, y) \mapsto (-x, -y)$ , which is the rule for the rotation of $180^\circ$ about the origin; in general $P^{'}$ is on the line $P O$ with $O P^{'} = O P$ , exactly where a half-turn about $O$ sends $P$ .

Sources and credits

Written by Ganzaa.
Video: Enlargements with Negative Scale Factors, Corbettmaths, embedded from YouTube.

The diagrams, worked examples and practice items were written by Ganzaa.

1An upside-down picture

2Finding the centre and the scale factor

An upside-down picture

Finding the centre and the scale factor