Enlargements using coordinates/digital geometry | Year 9

01 / LEARN

Understand the idea.

Scale from the centre

An enlargement scales distances from a fixed centre. For positive factor $k$ , the centre, original point and image lie on one ray, with the image distance equal to $k$ times the original distance.

From the origin, scale both coordinates: $(x, y) \mapsto (k x, k y)$ . A factor above $1$ enlarges; a factor between $0$ and $1$ reduces. Negative coordinates retain their signs when multiplied by a positive factor.

(x, y) \mapsto (k x, k y)

On an equal-scale one-unit coordinate grid, the sage original joins A at 1,1, B at 3,1 and C at 1,2. Its gold image joins A prime at 2,2, B prime at 6,2 and C prime at 2,4.

Use a different centre

For another centre $C$ , use the displacement from $C$ to the point $P$ . Subtract the centre coordinates, multiply that displacement by $k$ , then add the centre back: $P^{'} = C + k (P - C)$ .

Check that the original and image align with the centre and have the correct distance ratio. A digital plot can reveal an error, but the coordinate calculation records the centre and scale factor that actually define the enlargement.

P^{'} = C + k (P - C)

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Enlarge a triangle

Enlarge A(1,1), B(3,1), C(1,2) by factor 2 from the origin.

Worked solution

1
State the coordinate rule.
$(x, y) \mapsto (2 x, 2 y)$
The centre is the origin and the scale factor is $2$ .
2
Transform the first vertex.
$A (1, 1) \mapsto A^{'} (2, 2)$
Both coordinate displacements from the origin double.
3
Transform the remaining vertices.
$\begin{aligned} B (3, 1) & \mapsto B^{'} (6, 2) \\ C (1, 2) & \mapsto C^{'} (2, 4) \end{aligned}$
The same scale factor applies to every vertex.
4
Check two corresponding lengths.
$\begin{aligned} A B & = 2, & A^{'} B^{'} & = 4 \\ A C & = 1, & A^{'} C^{'} & = 2 \end{aligned}$
Both the horizontal and vertical sides have doubled, consistent with the enlargement.

Answer

A′(2,2), B′(6,2), C′(2,4).

EXAMPLE 02

A shifted centre

Enlarge P(3,2) by factor 3 from C(1,1).

Worked solution

1
Find the displacement from the centre.
$P - C = (3 - 1, 2 - 1) = (2, 1)$
The distance must be scaled relative to $C (1, 1)$ , rather than the origin.
2
Scale that displacement.
$3 (P - C) = 3 (2, 1) = (6, 3)$
An enlargement by factor $3$ triples each displacement component.
3
Add the centre back.
$P^{'} = (1, 1) + (6, 3) = (7, 4)$
Adding $C$ converts the scaled displacement into the image coordinates.
4
Check alignment and the scale factor.
$P^{'} - C = (6, 3) = 3 (2, 1)$
The image displacement is a positive triple of the original, so it lies on the same ray at three times the distance.

Answer

P′(7,4).

03 / YOUR TURN

Try it yourself.

Check your understanding

Enlarge (−4,6) by factor 1/2 from the origin.

Show answer & reasoning

(−2,3)

Multiply both coordinates by $\frac{1}{2}$ : $(\frac{1}{2} (- 4), \frac{1}{2} (6)) = (- 2, 3)$ . The positive factor halves both distances without reversing their signs.

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

Enlarge $P (2, - 3)$ by factor 2 from the origin.
2 marks
Enlarge $Q (- 6, 4)$ by factor $\frac{1}{2}$ from the origin.
2 marks
Enlarge $A (1, 0)$ , $B (3, 0)$ , $C (1, 2)$ by factor 3 from the origin.
3 marks
Centre $C$ is $(1, - 1)$ . Enlarge $P (3, 2)$ by factor 2.
3 marks
An enlargement from the origin maps $(2, 3)$ to $(6, 9)$ . Find $k$ .
2 marks
A student maps $(3, 1)$ to $(5, 3)$ and calls it factor 2 from the origin. Give the correct image and describe the student’s movement.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Enlarge $P (2, - 3)$ by factor 2 from the origin.
2 marks
$2 \times 2 = 4$ ; $2 \times (- 3) = - 6$ .
$P^{'} (4, - 6)$ .
Enlarge $Q (- 6, 4)$ by factor $\frac{1}{2}$ from the origin.
2 marks
Halve both coordinates.
$Q^{'} (- 3, 2)$ .
Enlarge $A (1, 0)$ , $B (3, 0)$ , $C (1, 2)$ by factor 3 from the origin.
3 marks
Multiply every coordinate by 3.
Both perpendicular sides triple.
$A^{'} (3, 0)$ , $B^{'} (9, 0)$ , $C^{'} (3, 6)$ .
Centre $C$ is $(1, - 1)$ . Enlarge $P (3, 2)$ by factor 2.
3 marks
$P - C = (2, 3)$ .
Double to $(4, 6)$ , then add $C$ to get $(5, 5)$ .
$P^{'} (5, 5)$ .
An enlargement from the origin maps $(2, 3)$ to $(6, 9)$ . Find $k$ .
2 marks
$6 \div 2 = 3$ and $9 \div 3 = 3$ . Both coordinates give the same factor.
$3$
A student maps $(3, 1)$ to $(5, 3)$ and calls it factor 2 from the origin. Give the correct image and describe the student’s movement.
3 marks
The correct image doubles both coordinates.
The student added the same displacement to the point.
Correct image $(6, 2)$ ; the student translated by $(2, 2)$ .

Checking saved progress…

Curriculum reference & lesson scope

Australian Curriculum v9 / South Australian Curriculum

Angle reasoning consolidation; enlargements using coordinates/digital geometry; invariant and changing properties; geometric constructions; sequences of steps and decisions; testing/refining construction algorithms; explaining geometric reasoning.

Official curriculum source ↗

In this lesson

Understand the idea.

Scale from the centre

Use a different centre

See the method in action.

Enlarge a triangle

State the coordinate rule.

Transform the first vertex.

Transform the remaining vertices.

Check two corresponding lengths.

A shifted centre

Find the displacement from the centre.

Scale that displacement.

Add the centre back.

Check alignment and the scale factor.

Try it yourself.

Check your understanding

Keep practising