Use positive fractional enlargement factors
Construct positive fractional enlargements and reductions, including genuinely free-centre tasks with more than one valid image and no hidden compulsory pivot.
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
- Use the stated-centre enlargement formula.
- Multiply signed fractions exactly.
- Explain a common centre and scale ratio on every ray.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which positive fractional factor makes a figure larger?
What you will learn
- Interpret positive factors below and above 1 without reversing rays.
- Choose and record a valid centre when none is supplied.
- Judge a free-centre image against its own centre and factor rather than one reference coordinate list.
Polygon ABC has A(1,1), B(4,1), C(2,3). In the given order, enlarge about (0,0) with positive factor 1/2. 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 0.5 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.
A fractional factor still acts on the entire radius
For 0<k<1, the image lies between the centre and each noncentral object point on the same ray. The figure becomes smaller, but its boundary order and angles remain unchanged. A noninteger factor greater than 1 instead moves the image farther out; a fraction is not automatically a reduction.
Keep the factor exact while scaling the signed local displacements. A negative coordinate in the final image does not mean the scale factor was negative: it may simply locate a point to the left or below the coordinate origin. The sign of k concerns the direction from the enlargement centre.
Choose a centre only when the question permits it
A task giving k without a centre leaves a genuine choice. Select a point C, label or record it explicitly, and construct every image vertex with P′=C+k(P−C). Use that same C for the whole figure. Moving the pivot separately for different vertices does not define one enlargement.
For k≠1, another centre generally produces a different but equally valid image. A worked reference can show one choice without making it compulsory. Validate the learner’s own centre, common factor and complete image; do not compare only its absolute coordinates with one privately chosen reference pivot.
Explain what varies and what stays the same
With the same positive factor, all valid free-centre images have the same corresponding length ratio and equal angles. Their positions can differ because the centre changes the common offset(1−k)C. Show that difference with a second deliberately chosen centre rather than calling one valid result a mistake.
If k=1, the offset vanishes and every centre produces exactly the original figure, so choosing a different centre cannot produce a different identity image. Keep this exception alongside the positive-factor restriction. Full labelled construction and practical accuracy remain work to demonstrate; a permissive free-centre choice does not waive the need for consistent rays and factor.
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
Two learners use the same k≠1 but choose different permitted centres. What must be checked?
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 1/2. 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 1/2. The scaled local displacement is (1/2,1/2); adding O gives A′=(1/2,1/2).
- Use P′=O+k(P−O) for every vertex. This choice gives A′=(1/2,1/2), B′=(2,1/2), C′=(1,3/2). The image stays between the centre and each noncentral point.
- Join the complete named image boundary and check each centre-to-image displacement is 1/2 times its object displacement. The inverse is enlargement about (0,0) with positive factor 2. Positive factors keep the same ray sense; do not infer a negative factor from negative global coordinates.
Answer A′=(1/2,1/2), B′=(2,1/2), C′=(1,3/2).
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
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 (1,2) with positive factor 2/3. Construct and label its image under the stated transformation; record the corresponding vertices and explain the method.
Give me a hint
Check whether the centre is given or genuinely free, then use one consistent pivot for all vertices.
Compare my reasoning
- The given centre is O=(1,2). For A, the local radius is (-1,-2) and is multiplied by positive factor 2/3. The scaled local displacement is (-2/3,-4/3); adding O gives A′=(1/3,2/3).
- Use P′=O+k(P−O) for every vertex. This choice gives A′=(1/3,2/3), B′=(7/3,2/3), C′=(3,2), D′=(1,8/3). The image stays between the centre and each noncentral point.
- Join the complete named image boundary and check each centre-to-image displacement is 2/3 times its object displacement. The inverse is enlargement about (1,2) with positive factor 3/2. Positive factors keep the same ray sense; do not infer a negative factor from negative global coordinates.
A′=(1/3,2/3), B′=(7/3,2/3), C′=(3,2), D′=(1,8/3).
Look for these in your work
- Interpret positive factors below and above 1 without reversing rays.
- Choose and record a valid centre when none is supplied.
- Judge a free-centre image against its own centre and factor rather than one reference coordinate list.
Polygon ABCDEF has A(0,0), B(4,0), C(4,1), D(1,1), E(1,3), F(0,3). Use positive scale factor 1/4; no centre has been specified. No centre is supplied. Choose and record your own centre, then construct an image at the stated positive scale factor. Explain why another centre could give a different valid image.
Give me a hint
Check whether the centre is given or genuinely free, then use one consistent pivot for all vertices.
Compare my reasoning
- No centre was supplied, so one permissible choice is O=(-2,-1). Record this chosen pivot explicitly. It is one permitted choice, and another consistent centre is also allowed. For A, the local radius is (2,1) and is multiplied by positive factor 1/4. The scaled local displacement is (1/2,1/4); adding O gives A′=(-3/2,-3/4).
- Use P′=O+k(P−O) for every vertex. This choice gives A′=(-3/2,-3/4), B′=(-1/2,-3/4), C′=(-1/2,-1/2), D′=(-5/4,-1/2), E′=(-5/4,0), F′=(-3/2,0). The image stays between the centre and each noncentral point.
- Join the complete named image boundary and check each centre-to-image displacement is 1/4 times its object displacement. The inverse is enlargement about (-2,-1) with positive factor 4. Positive factors keep the same ray sense; do not infer a negative factor from negative global coordinates.
- For a second valid choice O=(0,0), the image is A′=(0,0), B′=(1,0), C′=(1,1/4), D′=(1/4,1/4), E′=(1/4,3/4), F′=(0,3/4). These different absolute coordinates are also valid because they consistently use that different permitted centre and the same positive factor.
One valid choice: O=(-2,-1), k=1/4; A′=(-3/2,-3/4), B′=(-1/2,-3/4), C′=(-1/2,-1/2), D′=(-5/4,-1/2), E′=(-5/4,0), F′=(-3/2,0). Other explicitly chosen centres are valid when the whole image uses P′=O+k(P−O).
Look for these in your work
- Interpret positive factors below and above 1 without reversing rays.
- Choose and record a valid centre when none is supplied.
- Judge a free-centre image against its own centre and factor rather than one reference coordinate list.
Polygon ABCD has A(-2,1), B(1,1), C(2,4), D(-1,3). Use positive scale factor 3/2; no centre has been specified. No centre is supplied. Choose and record your own centre, then construct an image at the stated positive scale factor. Explain why another centre could give a different valid image.
Give me a hint
Check whether the centre is given or genuinely free, then use one consistent pivot for all vertices.
Compare my reasoning
- No centre was supplied, so one permissible choice is O=(-1,1). Record this chosen pivot explicitly. It is one permitted choice, and another consistent centre is also allowed. For A, the local radius is (-1,0) and is multiplied by positive factor 3/2. The scaled local displacement is (-3/2,0); adding O gives A′=(-5/2,1).
- Use P′=O+k(P−O) for every vertex. This choice gives A′=(-5/2,1), B′=(2,1), C′=(7/2,11/2), D′=(-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 (-1,1) with positive factor 2/3. Positive factors keep the same ray sense; do not infer a negative factor from negative global coordinates.
- For a second valid choice O=(0,0), the image is A′=(-3,3/2), B′=(3/2,3/2), C′=(3,6), D′=(-3/2,9/2). These different absolute coordinates are also valid because they consistently use that different permitted centre and the same positive factor.
One valid choice: O=(-1,1), k=3/2; A′=(-5/2,1), B′=(2,1), C′=(7/2,11/2), D′=(-1,4). Other explicitly chosen centres are valid when the whole image uses P′=O+k(P−O).
Look for these in your work
- Interpret positive factors below and above 1 without reversing rays.
- Choose and record a valid centre when none is supplied.
- Judge a free-centre image against its own centre and factor rather than one reference coordinate list.
Common mistakes
- Assuming every fraction is less than 1.
- Using different centres for different vertices.
- Treating one reference centre as compulsory when no centre was given.
What can you explain now?
No centre is given and k=2/5. Must a learner reproduce the reference centre?
Compare with the explanation
No. Any explicitly chosen centre is valid if the entire image uses that centre and the given factor consistently.
A fixed reference can illustrate one construction, but the original task leaves the pivot open.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain a positive reduction and a fractional enlargement, then describe how two permitted centres can give different valid images.
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.