Enlarge shapes about specified centres
Construct positive enlargements from a stated centre, using complete centre-to-vertex rays and preserving the factor 1 identity rather than multiplying global coordinates indiscriminately.
Exam scope and assessment notes
These twelve shared lessons apply to Foundation and inherited Higher for F5.2A–M; no extra Higher row is added. Keep named corners, signed turns, equivalent mirror equations, exact fractions, positive factors and the freely chosen-centre and identity exceptions. General-angle coordinates are final3dp approximations of unrounded geometry. Written explanations need human review; coordinate agreement does not certify proof or physical construction. Linear November2026/June2027 entry and regional-centre rules remain separate from modular awards. Book2 Shape and Space7 tags and Hubs66–71 are navigation only: map223–248 differs from proof222/summary247/next248, with chapter/Hub bodies unavailable.
Before you begin
- Find signed displacements from a given point.
- Multiply coordinates by positive integer and fractional factors.
- Track labelled rays and full polygon boundary order.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which formula enlarges P about a non-origin centre C?
What you will learn
- Scale each displacement from the stated centre by the same positive factor.
- Keep centre vertices fixed and label every image boundary point.
- Use reciprocal factors for inverse checks and explain the identity-centre ambiguity.
Polygon ABC has A(1,1), B(4,1), C(2,3). In the given order, enlarge about (0,0) with positive factor 2. Construct and label its image under the stated transformation; record the corresponding vertices and explain the method.
Describe your operation, construct every named image point, and record your actual directed traces. Only the question supplies the givens. The board does not transform a shape or recover a missing parameter for you.
Both Cartesian directions have equal drawing scale. Blue is supplied geometry, amber is my actual work, purple is dirty retained work, and green is a separately gated reference. Screen size is not physical instrument calibration.
Actual boundaries retain their entered order, closure, concavity and wrong or incomplete points. No reflected corner, rotation endpoint, mirror foot or target polygon is filled automatically. Construction rays continue beyond their anchors; mirror lines continue both ways; joining segments keep their original endpoints.
Supplied positive enlargement factor 2 about (0, 0). This description does not generate a target boundary.
This finite fixed-case comparison is not proof marking, marks, mastery, physical construction calibration or specialist/formal assessment approval. General-angle points stay unrounded; a separately requested three-decimal written table is an approximation.
Actual operation 1
No actual construction is recorded for this action.
Choose a point slot before adding a keyboard coordinate or picking on the board. Named picks retain exact given or current clean recorded coordinates. Free picks retain their unrounded actual planar coordinates; they do not snap to an expected result.
My written coordinates, quantities and reasoning
Write both coordinates as x,y; exact fractions and finite decimals are supported.
Write both coordinates as x,y; exact fractions and finite decimals are supported.
Write both coordinates as x,y; exact fractions and finite decimals are supported.
Explain your actual correspondence, construction and conditions. A written attempt is not an automatically verified proof.
Explain your actual correspondence, construction and conditions. A written attempt is not an automatically verified proof.
Explain your actual correspondence, construction and conditions. A written attempt is not an automatically verified proof.
This sequence is optional. If you add operations, complete every chosen parameter or remove the sequence. The tool generates no inverse for you.
These original cases support mathematical self-checking. They do not award marks, mastery, proof approval, physical construction competence, specialist signoff or formal assessment readiness. Positive scale factors only; unchanged images do not uniquely recover a centre. Keep general-angle construction coordinates unrounded and separately requested written coordinates at three decimal places.
Measure from the centre, not from the axes
For enlargement with centre C and factor k, the image satisfies CP′=kCP on the same ray because k is positive. In coordinates this is P′=C+k(P−C). First subtract the centre, scale the whole displacement and then restore the centre’s position.
Multiplying global coordinates by k only works when the centre is the origin. A non-origin pivot changes the offsets that must be scaled. Draw the centre and the rays to all vertices, and check that each image lies in the correct forward sense at the required multiple of its original radius.
Include every vertex and the identity factor
The centre itself stays fixed. If a polygon vertex coincides with it, that vertex and its image overlap legitimately. Scale every other vertex and reconnect the named points in the original boundary order, retaining a concave indentation rather than filling it in.
A factor greater than 1 moves noncentral points farther along their rays, while a factor between 0 and 1 moves them towards the centre. The positive factor 1 leaves every point unchanged. It is allowed here; zero would collapse the figure and negative factors would reverse rays, outside this section’s stated positive-factor scope.
Check the inverse and distinguish a given centre
For k>0, enlargement about the same centre by 1/k undoes the image. Use the local displacements to check both factors rather than scaling already rounded global coordinates. Corresponding lengths multiply by k and angles remain equal throughout the construction.
When a centre is explicitly supplied, keep it in the complete description even at k=1. However, if only an unchanged object/image pair were shown, the identity would not allow that centre to be recovered uniquely: every centre gives the same identity map. Supplied information and recoverable information are different claims.
Sources for this lesson
- Pearson Mathematics A linear specification · Issue2 · Transformation geometry Foundation printed23/physical29 and Higher inheritance printed38/physical44
- Pearson 2026/2027 International GCSE entry manual · centre regions physical3 and Mathematics A physical25
- Pearson literal second-edition book/Hub map · Book2 Shape and Space7 physical12 · navigation only
Pause and explain
Which positive scale factor undoes enlargement by 5/2 about C?
Worked example
Polygon ABC has A(1,1), B(4,1), C(2,3). In the given order, enlarge about (0,0) with positive factor 2. Construct and label its image under the stated transformation; record the corresponding vertices and explain the method.
Show the worked solution
- The given centre is O=(0,0). For A, the local radius is (1,1) and is multiplied by positive factor 2. The scaled local displacement is (2,2); adding O gives A′=(2,2).
- Use P′=O+k(P−O) for every vertex. This choice gives A′=(2,2), B′=(8,2), C′=(4,6). Noncentral image points lie farther along the same forward rays.
- Join the complete named image boundary and check each centre-to-image displacement is 2 times its object displacement. The inverse is enlargement about (0,0) with positive factor 1/2. Positive factors keep the same ray sense; do not infer a negative factor from negative global coordinates.
Answer A′=(2,2), B′=(8,2), C′=(4,6).
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Polygon ABCD has A(0,0), B(3,0), C(4,2), D(1,3). In the given order, enlarge about (1,-1) with positive factor 3. Construct and label its image under the stated transformation; record the corresponding vertices and explain the method.
Give me a hint
Scale the whole displacement from the supplied centre and keep the positive ray sense.
Compare my reasoning
- The given centre is O=(1,-1). For A, the local radius is (-1,1) and is multiplied by positive factor 3. The scaled local displacement is (-3,3); adding O gives A′=(-2,2).
- Use P′=O+k(P−O) for every vertex. This choice gives A′=(-2,2), B′=(7,2), C′=(10,8), D′=(1,11). Noncentral image points lie farther along the same forward rays.
- Join the complete named image boundary and check each centre-to-image displacement is 3 times its object displacement. The inverse is enlargement about (1,-1) with positive factor 1/3. Positive factors keep the same ray sense; do not infer a negative factor from negative global coordinates.
A′=(-2,2), B′=(7,2), C′=(10,8), D′=(1,11).
Look for these in your work
- Scale each displacement from the stated centre by the same positive factor.
- Keep centre vertices fixed and label every image boundary point.
- Use reciprocal factors for inverse checks and explain the identity-centre ambiguity.
Polygon ABCDEF has A(0,0), B(4,0), C(4,1), D(1,1), E(1,3), F(0,3). In the given order, enlarge about (-2,1) with positive factor 3/2. Construct and label its image under the stated transformation; record the corresponding vertices and explain the method.
Give me a hint
Scale the whole displacement from the supplied centre and keep the positive ray sense.
Compare my reasoning
- The given centre is O=(-2,1). For A, the local radius is (2,-1) and is multiplied by positive factor 3/2. The scaled local displacement is (3,-3/2); adding O gives A′=(1,-1/2).
- Use P′=O+k(P−O) for every vertex. This choice gives A′=(1,-1/2), B′=(7,-1/2), C′=(7,1), D′=(5/2,1), E′=(5/2,4), F′=(1,4). Noncentral image points lie farther along the same forward rays.
- Join the complete named image boundary and check each centre-to-image displacement is 3/2 times its object displacement. The inverse is enlargement about (-2,1) with positive factor 2/3. Positive factors keep the same ray sense; do not infer a negative factor from negative global coordinates.
A′=(1,-1/2), B′=(7,-1/2), C′=(7,1), D′=(5/2,1), E′=(5/2,4), F′=(1,4).
Look for these in your work
- Scale each displacement from the stated centre by the same positive factor.
- Keep centre vertices fixed and label every image boundary point.
- Use reciprocal factors for inverse checks and explain the identity-centre ambiguity.
Polygon ABCD has A(-2,1), B(1,1), C(2,4), D(-1,3). In the given order, enlarge about (3,4) with positive factor 1. Construct and label its image under the stated transformation; record the corresponding vertices and explain the method.
Give me a hint
Scale the whole displacement from the supplied centre and keep the positive ray sense.
Compare my reasoning
- The given centre is O=(3,4). For A, the local radius is (-5,-3) and is multiplied by positive factor 1. The scaled local displacement is (-5,-3); adding O gives A′=(-2,1).
- Use P′=O+k(P−O) for every vertex. This choice gives A′=(-2,1), B′=(1,1), C′=(2,4), D′=(-1,3). At factor 1 all vertices stay fixed.
- Join the complete named image boundary and check each centre-to-image displacement is 1 times its object displacement. The inverse is enlargement about (3,4) with positive factor 1. A supplied vertex at the centre stays fixed; the identity image alone cannot recover a unique centre.
A′=(-2,1), B′=(1,1), C′=(2,4), D′=(-1,3). This is the identity about the stated centre; the same image alone would also be compatible with any other centre.
Look for these in your work
- Scale each displacement from the stated centre by the same positive factor.
- Keep centre vertices fixed and label every image boundary point.
- Use reciprocal factors for inverse checks and explain the identity-centre ambiguity.
Common mistakes
- Scaling global coordinates when the centre is not the origin.
- Reversing a ray for a positive fractional factor.
- Inventing a uniquely recoverable centre from an identity image.
What can you explain now?
For positive factor 1, does the image reveal which centre was used?
Compare with the explanation
No. The same identity image is produced about every centre.
A stated centre can still be part of the supplied description, but it cannot be recovered uniquely from an unchanged image alone.
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Make it stick
Derive P′=C+k(P−C), explain all positive-factor ranges and state the inverse and identity caveats.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.