Enlargement and Similarity in 2D

Lessons

Senior 3 · Mathematics

Enlargement and Similarity in 2D

5 lessons

Lesson 2 of 5

Enlargement with a positive scale factor

25 min

A copying machine made of rods

Long before photocopiers, draughtsmen copied maps and drawings larger or smaller with a pantograph: a jointed frame of rods fixed at one point. A pointer traces the original, and a pencil further along the frame draws the copy. The fixed point, the pointer and the pencil always stay in one straight line, and the pencil is always the same number of times as far from the fixed point as the pointer is. That is exactly the rule of an enlargement.

Definition

An enlargement with centre $O$ and scale factor $k$ maps each point $P$ to the point $P^{'}$ (its image) such that

\overrightarrow{OP'} = k\,\overrightarrow{OP} .

For $k > 0$ , $P^{'}$ lies on the ray from $O$ through $P$ , and $OP' = k \times OP$ . The original shape is the object.

$k > 1$ : the image is larger than the object.
$0 < k < 1$ : the image is smaller (it is still called an enlargement).
$k = 1$ : every point stays where it is.

Example: with centre $O$ and $k = 3$ , a point 2 cm from $O$ moves to the point 6 cm from $O$ on the same ray.

Important

Properties of an enlargement with scale factor $k$ :

the image is similar to the object, with ratio $k$ : every length is multiplied by $k$ ;
angles are unchanged, and each side of the image is parallel to the matching side of the object;
the centre is the only point that does not move (when $k \neq 1$ ), and every line through the centre is mapped onto itself;
areas are multiplied by $k^{2}$ (lesson 5 explains why).

In coordinates. With centre the origin: $(x, y) \mapsto (kx, ky)$ . With centre $X (a, b)$ :

(x, y) \mapsto \big(a + k(x - a),\; b + k(y - b)\big), \quad\text{that is,}\quad \overrightarrow{XP'} = k\,\overrightarrow{XP}.

Worked example 1: enlarging from the origin

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

Solution Multiply each coordinate by 3: $A^{'} (3, 3)$ , $B^{'} (9, 3)$ , $C^{'} (3, 6)$ .

By construction: draw a ray from $O$ through each vertex, measure the distance from $O$ to the vertex, and mark the image three times as far along the same ray. Check: $A B = 2$ and $A'B' = 6 = 3 \times 2$ , and $A'B' \parallel AB$ .

Worked example 2: a centre that is not the origin

Enlarge triangle $P (2, 3)$ , $Q (4, 3)$ , $R (2, 5)$ with centre $X (1, 2)$ and scale factor 2.

Solution Use $\overrightarrow{XP'} = 2\,\overrightarrow{XP}$ for each vertex.

Vertex	$\overrightarrow{XP}$	$2\,\overrightarrow{XP}$	Image
$P (2, 3)$	$\begin{pmatrix} 1 \\ 1 \end{pmatrix}$	$\begin{pmatrix} 2 \\ 2 \end{pmatrix}$	$P^{'} (3, 4)$
$Q (4, 3)$	$\begin{pmatrix} 3 \\ 1 \end{pmatrix}$	$\begin{pmatrix} 6 \\ 2 \end{pmatrix}$	$Q^{'} (7, 4)$
$R (2, 5)$	$\begin{pmatrix} 1 \\ 3 \end{pmatrix}$	$\begin{pmatrix} 2 \\ 6 \end{pmatrix}$	$R^{'} (3, 8)$

Doubling the coordinates instead, $(4, 6)$ for $P$ , would be an enlargement about the origin, a different transformation.

Worked example 3: a fractional scale factor

Enlarge the square $K (2, 1)$ , $L (6, 1)$ , $M (6, 5)$ , $N (2, 5)$ with centre $X (8, 1)$ and scale factor $\frac{1}{2}$ .

Solution $\overrightarrow{XK} = \begin{pmatrix} -6 \\ 0 \end{pmatrix}$ , so $\overrightarrow{XK'} = \begin{pmatrix} -3 \\ 0 \end{pmatrix}$ and $K^{'} (5, 1)$ . In the same way $L^{'} (7, 1)$ , $M^{'} (7, 3)$ and $N^{'} (5, 3)$ .

The image is a square of side 2, half the side of the object, and each image vertex is halfway between $X$ and the object vertex.

Worked example 4: forwards and backwards

An enlargement has centre $X (2, - 1)$ and scale factor 4. (a) Find the image of $(3, 1)$ . (b) Which point has image $(10, 3)$ ?

Solution (a) $\overrightarrow{XP} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ , so $\overrightarrow{XP'} = \begin{pmatrix} 4 \\ 8 \end{pmatrix}$ and the image is $(2 + 4, - 1 + 8) = (6, 7)$ .

(b) $\overrightarrow{XP'} = \begin{pmatrix} 8 \\ 4 \end{pmatrix}$ , and $\overrightarrow{XP'} = 4\,\overrightarrow{XP}$ , so $\overrightarrow{XP} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ and the point is $(2 + 2, - 1 + 1) = (4, 0)$ .

Warning

Multiply distances from the centre, not from the object. With $k = 3$ , the image is 3 times as far from the centre as the object, not 3 units further on.

$(x, y) \mapsto (kx, ky)$ is only for the origin. For any other centre, work with the vector from the centre.

Tip

On squared paper: count the squares across and up from the centre to a vertex, multiply both counts by $k$ , and count again from the centre. That is $\overrightarrow{XP'} = k\,\overrightarrow{XP}$ done with your finger.

Corbettmaths: Enlargement with fractional scale factors. Watch how a scale factor such as one half moves every vertex towards the centre, as in Worked example 3.

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Practice

The last two are harder.

Enlarge the triangle $(1, 2)$ , $(3, 2)$ , $(1, 5)$ with centre the origin and scale factor 2.
Find the images of $(4, 6)$ and $(8, 2)$ under the enlargement with centre the origin and scale factor $\frac{1}{2}$ .
Find the image of $(2, 3)$ under the enlargement with centre $(1, 1)$ and scale factor 3.
Find the image of $(6, 4)$ under the enlargement with centre $(2, 0)$ and scale factor 1.5.
A triangle with sides 3 cm, 4 cm and 5 cm is enlarged with scale factor 2.5. Give the sides of the image. Is the image right-angled?
An enlargement has centre $(0, 4)$ and scale factor 2. Which point has image $(6, 10)$ ?
An enlargement with centre the origin maps $A (2, 3)$ to $A^{'} (5, 7.5)$ . Find the scale factor and the image of $B (- 4, 2)$ .
$A^{'}$ and $B^{'}$ are the images of $A$ and $B$ under an enlargement with centre $O$ and scale factor $k$ . Explain why $A'B' = k \times AB$ and $A'B' \parallel AB$ .

Show answersHide answers

$(2, 4)$ , $(6, 4)$ , $(2, 10)$ .
$(2, 3)$ and $(4, 1)$ .
$(1 + 3 \times 1, 1 + 3 \times 2) = (4, 7)$ .
$(2 + 1.5 \times 4, 0 + 1.5 \times 4) = (8, 6)$ .
7.5 cm, 10 cm and 12.5 cm. Yes: angles do not change, so the right angle stays a right angle.
$\overrightarrow{XP'} = \begin{pmatrix} 6 \\ 6 \end{pmatrix}$ , so $\overrightarrow{XP} = \begin{pmatrix} 3 \\ 3 \end{pmatrix}$ and the point is $(3, 7)$ .
$k = \frac{5}{2} = \frac{7.5}{3} = 2.5$ ; $B^{'} (- 10, 5)$ .
$\overrightarrow{A'B'} = \overrightarrow{OB'} - \overrightarrow{OA'} = k\,\overrightarrow{OB} - k\,\overrightarrow{OA} = k\,\overrightarrow{AB}$ . A vector that is $k$ times another is parallel to it and $k$ times as long (for $k > 0$ ).

Sources and credits

Written by Ganzaa.
Photo: Pantogr.jpg, Stefan Kühn, Wikimedia Commons, CC BY-SA 3.0.
Video: Enlargement with fractional scale factors, Corbettmaths, embedded from YouTube.

The diagrams, worked examples and practice items were written by Ganzaa. This lesson uses CC BY-SA material and is shared under CC BY-SA 4.0.

1A copying machine made of rods

A copying machine made of rods