Enlargement

Module 4 · Lesson 4.06 · Spec G7 · ⭐ High Priority

Positive, fractional and negative scale factors — and why the centre of enlargement is never optional.

The Fourth Transformation

In 4.05 we met three transformations that preserve size: rotation, reflection and translation. Enlargement is the fourth — and the only one that changes size. It still preserves shape, so the image is similar to the object (next lesson, 4.07).

Two required parameters

Scale factor (how much bigger or smaller) AND centre of enlargement (where it pivots from). Miss either → description is incomplete.

Three types of scale factor

Positive > 1 (larger), 0 < SF < 1 (smaller), negative (opposite side + rotated 180°). All three appear on GCSE papers.

AQA Evidence

Nov 2024 Paper 1 Q14 included enlargement by a negative fraction — singled out as a topic students struggled with. Negative and fractional scale factors are the hardest variants; we'll cover both in detail.

Enlargement: Two Required Parameters

An enlargement description must include both of these or the description mark is lost.

ParameterWhat it tells youExample

Scale factor (SF)How much bigger or smaller the image is compared to the objectSF = 3 → image lengths are 3× original lengths

Centre of enlargementThe fixed point from which every other point expands or contractsUsually a coordinate: (0, 0) or (2, 3)

Full description: Enlargement, scale factor k, centre (a, b)

Accepted wording examples

"Enlargement, scale factor 3, centre (0, 0)"

"Enlargement by scale factor ½, centre (2, 1)"

"Enlargement, scale factor −2, centre (0, 0)"

SF Notation Options

AQA accepts the scale factor in any sensible form: fraction (½), decimal (0.5), or ratio-style (1 : 2 where the first value corresponds to the object). Pick the form that matches how the question is asked.

AQA Mark Scheme

"Enlargement SF 3" alone = 0 marks (missing centre). "Enlargement centre (0, 0)" alone = 0 marks (missing SF). Both required, every time.

Positive Scale Factor > 1: Larger Image

If SF > 1, every length is multiplied by SF. The image is larger, on the same side of the centre as the original, in the same orientation.

Key rules for SF > 1

All image side lengths = SF × object side lengths.

Image is on the same side of the centre as the object.

Each image vertex lies on the ray from the centre through the corresponding object vertex, at distance SF × (object-to-centre distance).

Orientation preserved (not rotated, not flipped).

Fractional Scale Factor (0 < SF < 1): Smaller Image

If SF is a fraction between 0 and 1, the image is smaller than the object — but still on the same side of the centre, same orientation.

Key rules for 0 < SF < 1

Image side lengths = SF × object lengths (smaller because SF < 1).

Image between the centre and the object — same side, closer to the centre.

Orientation preserved. Shape still similar.

SF = ½ means the image is half as wide and half as tall — a quarter of the area (length ratios square).

Language Note

GCSE calls this "enlargement" even though the image is smaller. Do not call it a "reduction" or "shrink" — AQA won't award the mark for those words.

Negative Scale Factor: Opposite Side + Rotated 180°

A negative scale factor means two things happen at once: the image is scaled by |SF|, and it's placed on the opposite side of the centre (equivalent to also rotating 180° about the centre).

Key rules for negative SF

Image side lengths = |SF| × object lengths (so SF = −2 means 2× larger).

Image on the opposite side of the centre from the object.

The image is the object rotated 180° about the centre, then scaled by |SF|. (Orientation around the shape is preserved — like a rotation, not a reflection.)

Each image vertex is at distance |SF| × (object-to-centre distance) on the extended ray through the centre.

Special Case: SF = −1

An enlargement with scale factor −1 is equivalent to a 180° rotation about the centre. Same size, opposite side. This is the only SF where "enlargement" produces a congruent image.

Don't Call It Two Transformations

A negative-SF enlargement is a SINGLE transformation, not "enlargement then rotation". The description is simply "Enlargement, scale factor −2, centre (0, 0)".

Centre Not at the Origin

The centre of enlargement doesn't have to be (0, 0). When it isn't, every point's distance from the centre still scales by SF — just measured from the given centre.

Image of point P = Centre + SF × (P − Centre)

Step-by-Step

For each vertex P of the object, compute (P − centre) to get the displacement vector.

Multiply the displacement by SF.

Add the centre back to get the image point.

Repeat for every vertex, then join them up.

Ray Check

Draw a straight line (ray) from the centre through each object vertex. The corresponding image vertex must lie on the same ray (or its extension through the centre for negative SF). If not, you've made a mistake.

Finding the Centre of Enlargement

Given an object and its image, the centre can always be recovered.

Method — Ray Intersection

Pick a vertex on the object and its corresponding vertex on the image.

Draw a straight line through the two points, extended far enough to intersect the other rays.

Repeat for a second pair of corresponding vertices.

The two lines meet at the centre of enlargement.

A third pair is a useful check — it should pass through the same point.

Why It Works

Every image point lies on the ray from the centre through its corresponding object point. So the rays all pass through one shared point — the centre. Two rays are enough to locate it; a third confirms.

Coordinate Method (when centre is given)

If the centre is (a, b): object P = (x, y), image Q = (x', y'). The scale factor is k = (x' − a) / (x − a) = (y' − b) / (y − b). Both ratios must give the same k — that's a useful check.

Worked Example — Describe Fully (Positive SF)01

Triangle P has vertices at (2, 2), (4, 2), (2, 4). Triangle Q has vertices at (6, 6), (12, 6), (6, 12). Describe fully the single transformation that maps P onto Q. [3 marks]

M

Method: Same orientation + larger size ⇒ enlargement with positive SF > 1. Find SF from a pair of corresponding distances to a candidate centre. Try the origin first. SF × (object − centre) = image − centre.

C

Calculate: (2, 2) → (6, 6): 6/2 = 3 for x, 6/2 = 3 for y. (4, 2) → (12, 6): 12/4 = 3, 6/2 = 3 ✓. (2, 4) → (6, 12): 6/2 = 3, 12/4 = 3 ✓. Consistent SF = 3 from the origin, so centre = (0, 0).

E

Evaluate / Answer:Enlargement, scale factor 3, centre (0, 0). All three parameters present: type, SF, centre.

Mark split: B1 "enlargement"; B1 "scale factor 3"; B1 "centre (0, 0)". Any one missing = 0 for that mark.

Worked Example — Fractional Scale Factor02

Triangle X has vertices at (4, 4), (8, 4), (4, 8). Triangle Y has vertices at (2, 2), (4, 2), (2, 4). Describe fully the transformation X → Y. [3 marks]

M

Method: Smaller + same orientation ⇒ enlargement with fractional positive SF. Test SF = (image − centre) / (object − centre).

C

Calculate: Try centre (0, 0): (4, 4) → (2, 2), ratio 2/4 = ½. (8, 4) → (4, 2), ratio 4/8 = ½ and 2/4 = ½ ✓. (4, 8) → (2, 4), ratio 2/4 = ½ and 4/8 = ½ ✓. Consistent SF = ½ from origin.

E

Evaluate / Answer:Enlargement, scale factor ½, centre (0, 0).

Mark split: B1 "enlargement"; B1 "scale factor ½" (or 0.5, or 1/2); B1 "centre (0, 0)". Writing "scale factor 2" (inverse) is a common wrong answer — lose 1 mark.

Worked Example — Negative Scale Factor03

Triangle S has vertices at (1, 2), (2, 2), (1, 4). Enlarge S by scale factor −2, centre (0, 0). State the coordinates of the image. [3 marks]

M

Method: Apply image = centre + SF × (object − centre). With centre (0, 0) and SF = −2, this simplifies to image = (−2x, −2y).

C

Calculate: (1, 2) → (−2, −4). (2, 2) → (−4, −4). (1, 4) → (−2, −8).

E

Evaluate / Answer: image vertices are (−2, −4), (−4, −4), (−2, −8). Check: image is 2× the size of S, on the opposite side of the origin (it's the object rotated 180° about (0, 0), then scaled by 2).

Mark split: M1 for applying (x, y) → (−2x, −2y) or equivalent method; A1 A1 for correct coordinate pairs (one slip loses one A1).

Worked Example — Find Centre & SF04

Triangle R has vertices (2, 2), (3, 2), (2, 3). Triangle T has vertices (3, 3), (5, 3), (3, 5). Describe fully the transformation R → T. [3 marks]

M

Method: Same orientation + larger ⇒ positive enlargement, SF > 1. Test centre (0, 0) first; if ratios aren't consistent, find the centre using ray intersections.

C

Calculate: Centre (0, 0) test: (2, 2) → (3, 3) would need ratio 3/2 = 1.5; but (3, 2) → (5, 3) needs 5/3 ≈ 1.67 for x — not equal. So centre is NOT origin. Try centre (1, 1): (2,2) − (1,1) = (1,1); (3,3) − (1,1) = (2,2); ratio 2/1 = 2 ✓. (3,2) − (1,1) = (2,1); (5,3) − (1,1) = (4,2); ratios 4/2 = 2, 2/1 = 2 ✓. Consistent SF = 2 from centre (1, 1).

E

Evaluate / Answer:Enlargement, scale factor 2, centre (1, 1).

Mark split: B1 "enlargement scale factor 2"; B1 centre (1, 1); B1 all three parameters cohering. Most students lose the centre mark — this is the hardest part of enlargement questions. Drawing rays from corresponding vertices (extended) will show them meeting at (1, 1).

What Enlargement Changes & Preserves

Enlargement is the only GCSE transformation that changes size. This has consequences:

PropertyAfter enlargement

Size (lengths, area, perimeter)Changed — lengths × |SF|, area × SF², perimeter × |SF|

AnglesPreserved — every angle is identical

Parallel linesPreserved — parallel lines stay parallel

Shape / proportionsPreserved — object and image are similar (4.07)

Orientation (for positive SF)Preserved — not flipped, not rotated

Orientation (for negative SF)Rotated 180° about the centre

Fixed pointsOnly the centre of enlargement

Preview of 4.07 / 4.08

Because angles are preserved and sides are in a fixed ratio, object and image are similar shapes. This is the foundation for congruence & similarity (4.07) and area / volume scale factors (4.08, Higher).

All Four Transformations Compared

With enlargement, you now have the complete GCSE transformation toolkit.

TransformationParametersPreserves size?

RotationAngle + direction + centreYes — congruent image

ReflectionMirror lineYes — congruent image

TranslationColumn vectorYes — congruent image

EnlargementScale factor + centreNo — similar image (unless SF = ±1)

Identification checklist

Are the shapes the same size? Yes → rotation, reflection, or translation. No → enlargement.

Orientation flipped? → reflection.

Every vertex shifted by same vector? → translation.

Orientation preserved but different position? → rotation.

If sizes differ: check SF and centre → enlargement.

Strategy: "Describe Fully" Questions

Transformation questions worth 2–4 marks almost always include "describe fully". Use this fixed routine:

Step 1 — Compare sizes

Measure or compute a pair of corresponding sides. If same size ⇒ rotation / reflection / translation. Different size ⇒ enlargement.

Step 2 — If enlargement: find SF

SF = image length ÷ corresponding object length. Check the sign: image on same side of centre ⇒ positive SF. Image on opposite side (upside down) ⇒ negative SF.

Step 3 — Find the centre

Draw rays through pairs of corresponding vertices. They meet at the centre. Or test common centres (origin first) with the ratio check.

Step 4 — Write both parts

"Enlargement, scale factor k, centre (a, b)" — exactly that format. Three pieces of information: type, SF, centre.

Single Transformation

If "single transformation" is stated, never answer "enlargement then translation". A single enlargement with a non-origin centre already does both — that's why the centre matters.

Examiner Alerts

Alert 1 — Missing centre

"Enlargement, scale factor 3" without a centre scores 0 for the description. EVERY enlargement description needs both the scale factor AND the centre, no exceptions.

Alert 2 — Calling it a "reduction"

When SF < 1 the image is smaller, but GCSE terminology is still "enlargement". "Reduction" / "shrink" / "decrease" are NOT accepted answers.

Alert 3 — Negative SF misinterpreted

SF = −2 means the image is on the opposite side of the centre and 2× larger (NOT smaller, NOT a "mirror"). Students commonly write "scale factor 2, rotated 180°" as two separate transformations — that's wrong when the question says "single".

Alert 4 — Inverted scale factor

When image is smaller than object, some students write SF = 2 (thinking "object is twice the image"). But SF is defined as image/object, so if image is half the size, SF = ½.

Alert 5 — Centre assumed to be origin

Always check the centre — don't assume (0, 0). Test the ratio (image − centre) / (object − centre) for two pairs; if they don't agree, the centre isn't where you assumed.

Summary

Full description (always needed)

Enlargement, scale factor k, centre (a, b) — three pieces of information.

Missing any one → description mark lost.

Scale factor rules

SF > 1: larger image, same side, same orientation.

0 < SF < 1: smaller image, same side, same orientation.

SF = 1: no change (identity).

SF < 0 (negative): on the opposite side of the centre; |SF|× size; rotated 180°.

Key formula

Image of P = Centre + SF × (P − Centre).

When centre is the origin: image = SF × P.

Finding the centre

Draw rays through pairs of corresponding vertices; they meet at the centre.

Check with a third pair.

Never Forget

Scale factor AND centre. Every single enlargement description. Every single time.

Checkpoint

You should now be able to

Perform an enlargement given a scale factor and centre, including fractional and negative SFs.

Describe an enlargement fully with type, scale factor, and centre.

Find the centre of enlargement from an object and its image.

Interpret negative SFs as image on the opposite side of the centre.

Distinguish enlargement from the three size-preserving transformations (4.05).

Test yourself in under two minutes

1

Triangle A (1, 0), (3, 0), (1, 2) maps to triangle B (2, 0), (6, 0), (2, 4). Describe fully.

2

Enlarge triangle with vertices (0, 0), (2, 0), (0, 2) by SF = −1 centre (0, 0). State image coordinates.

3

An enlargement has SF = 3 centre (2, 1). The object has a vertex at (4, 3). Where's the image vertex?

4

A SF = ½ enlargement scales area by what factor?

Answers: 1) Ratios 2/1 = 2, 6/3 = 2, 4/2 = 2, all from origin ⇒ enlargement SF 2 centre (0, 0). 2) Each (x, y) → (−x, −y): (0, 0), (−2, 0), (0, −2) (same as 180° rotation about origin when SF=−1). 3) Centre + SF×(vertex − centre) = (2, 1) + 3×(2, 2) = (8, 7). 4) Area × (½)² = ¼ (area scale factor = SF², covered more in 4.08 Higher).

Next Lesson (4.07)

We build directly on enlargement by studying congruence and similarity — when two shapes are "the same" (congruent) vs "proportional" (similar), and how to use length ratios in problems.