Determining Centre of Enlargement and Linear Scale Factor

By the end of this lesson, learners should be able to

1. Locate the centre of enlargement of two similar figures by constructing ray lines through corresponding vertices.
2. Calculate the linear scale factor of an enlargement using corresponding side lengths.
3. Verify the linear scale factor using the ratio of distances from the centre of enlargement to image and object points.

Key vocabulary

Centre of enlargement

The fixed point where straight lines connecting all matching object and image vertices intersect.

Rays connecting corresponding vertices intersect at $(1,1)$, the centre of enlargement.

Linear scale factor

The ratio of an image length to the corresponding length on the original object shape.

Ratio of image length $A^{\mathrm{\prime }}{B}^{\mathrm{\prime }}=4$ to object length $AB=2$ gives scale factor $2$.

Corresponding vertices

Matching corner points on an object figure and its enlarged image shape.

Point $L(4,2)$ on triangle $LMN$ corresponds to vertex $L^{\mathrm{\prime }}(7,3.5)$ on triangle $L^{\mathrm{\prime }}{M}^{\mathrm{\prime }}{N}^{\mathrm{\prime }}$.

Lesson preview

Slide 1 · Finding the Centre of Enlargement and Scale FactorOpening · 18 min

When a figure undergoes enlargement, straight lines joining matching pairs of vertices intersect at a unique common location known as the centre of enlargement. Connecting matching vertices like $A$ to $A^{\mathrm{\prime }}$ and $B$ to $B^{\mathrm{\prime }}$ reveals this intersection. The linear scale factor is the ratio of any image side length to its corresponding object side length, or the ratio of distance from the centre of enlargement to an image point over the distance to the matching object point. Use Figure 1 to see how connecting corresponding vertices reveals the centre of enlargement.

• Lines passing through matching object and image vertices intersect at the unique centre of enlargement.
• The linear scale factor compares the length of an image side to its corresponding object side.
• Distances measured from the centre of enlargement to matching vertices share the exact same scalar ratio.
• Scale factor $k$ equals the image distance divided by object distance from the centre of enlargement.

Worked example. Draw straight lines through corresponding vertices: line $A{A}^{\mathrm{\prime }}$ through $(2,3)$ and $(3,5)$, line $B{B}^{\mathrm{\prime }}$ through $(4,3)$ and $(7,5)$, and line $C{C}^{\mathrm{\prime }}$ through $(2,6)$ and $(3,11)$. Find the common point of intersection: line $A{A}^{\mathrm{\prime }}$ has equation $y=2x-1$, line $B{B}^{\mathrm{\prime }}$ has equation $y=\frac{2}{3}x+\frac{1}{3}$, and both meet at $(1,1)$. Thus, the centre of enlargement is $O(1,1)$. Calculate object length $AB=4-2=2$ units and image length $A^{\mathrm{\prime }}{B}^{\mathrm{\prime }}=7-3=4$ units. Compute the linear scale factor $k=\frac{A^{\mathrm{\prime }}{B}^{\mathrm{\prime }}}{AB}=\frac{4}{2}=2$. Answer: Centre of enlargement is $(1,1)$ and linear scale factor $k=2$..

Check question: A quadrilateral $DEFG$ with side $DE=3\text{ cm}$ is enlarged to $D^{\mathrm{\prime }}{E}^{\mathrm{\prime }}{F}^{\mathrm{\prime }}{G}^{\mathrm{\prime }}$ with side $D^{\mathrm{\prime }}{E}^{\mathrm{\prime }}=9\text{ cm}$. Line $D{D}^{\mathrm{\prime }}$ passes through $(0,2)$ and $(2,6)$, while line $E{E}^{\mathrm{\prime }}$ passes through $(4,2)$ and $(2,6)$. What are the centre of enlargement and the linear scale factor?

1. Centre of enlargement $(2,6)$ and linear scale factor $3$
2. Centre of enlargement $(2,6)$ and linear scale factor $\frac{1}{3}$
3. Centre of enlargement $(0,0)$ and linear scale factor $3$
4. Centre of enlargement $(2,4)$ and linear scale factor $6$

Show answer

Centre of enlargement $(2,6)$ and linear scale factor $3$. The centre of enlargement is the point where the lines passing through corresponding vertices intersect. Both line $D{D}^{\mathrm{\prime }}$ and line $E{E}^{\mathrm{\prime }}$ pass through $(2,6)$, so this must be the centre of enlargement. The linear scale factor is the ratio of the image length to the corresponding object length: $k=\frac{D^{\mathrm{\prime }}{E}^{\mathrm{\prime }}}{DE}=\frac{9}{3}=3$.

Slide 2 · Verifying Scale Factors Using Distances from the CentreModel · 22 min

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• Both slides with diagrams and class activities
• Teacher notes, SNE adaptations and assessment checks
• Homework and the other 10 lessons in this sub-strand

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All 11 lessons in Similarity and Enlargement

1. 1Determining Centre of Enlargement and Linear Scale FactorDetermine the centre of enlargement and the linear scale factor for similar figures
2. 2Finding the Central Point of Dilation and Scale FactorsDetermine the centre of enlargement and the linear scale factor for similar figures
3. 3Constructing Images under Enlargement with Positive and Negative Scale FactorsConstruct the image of an object under an enlargement given the centre of enlargement and the linear scale factor (L.S.F)
4. 4Constructing Images under Geometric EnlargementConstruct the image of an object under an enlargement given the centre of enlargement and the linear scale factor (L.S.F)
5. 5Calculating Area and Volume Scale Factors for Similar FiguresDetermine the area and volume scale factor of different figures and solids
6. 6Determining Area and Volume Scale FactorsDetermine the area and volume scale factor of different figures and solids
7. 7Connecting Length, Surface Area, and Volume RatiosRelate linear scale factor, area scale factor, and volume scale factor in enlargements
8. 8Connecting Linear, Surface Area, and Volumetric Scale Factors in Similar SolidsRelate linear scale factor, area scale factor, and volume scale factor in enlargements
9. 9Applications of Similarity and Enlargement in Real LifeApply similarity and enlargement to real-life situations
10. 10Applying Similarity and Enlargement in Real-Life ContextsApply similarity and enlargement to real-life situations
11. 11Appreciating Similarity and Enlargement in Real-Life ApplicationsAppreciate the use of similarity and enlargement in real-life situations