eduKatePunggol · Secondary 4 Mathematics
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Enlargement From a Centre Other Than the Origin. Choose the reading route closest to your question.
Route 1 · Chapters 1–2
Route 2 · Chapters 3–4
Route 3 · Chapters 5–6
Route 4 · Chapters 7–8
Full chapter index · Independent practice · Secondary 4 Mathematics support
If your child doubles every coordinate when enlarging from C = (2, 1), check the displacement from C first. For P = (5, 3), the displacement is (3, 2). A scale factor of two makes it (6, 4), so the image is P′ = (8, 5). Directly doubling P to (10, 6) would use the origin as the centre.
Secondary 4 Mathematics tuition in Punggol can connect coordinates to the geometry: P′ = C + k(P − C). A Punggol Secondary 4 Maths tutor should ask which point stays fixed and which distance is being scaled. The scale factor applies to displacement from the stated centre, not automatically to coordinates measured from the origin.
This guide uses original coordinate examples, positive and negative scale factors and short independent checks. Topic timing and permitted methods can differ, so match practice to the student’s current work. Small-group 3-pax support should help learners plot and explain each image rather than apply coordinate multiplication without a reference point.
Full chapter index
Locate the centre · Chapters 1–2
Enlarge every vertex · Chapters 3–4
Check factors and inverses · Chapters 5–6
Practise and continue · Chapters 7–8
The centre of enlargement is the fixed reference point from which each object’s displacement is scaled. If C is the centre and P an original point, the directed displacement CP is multiplied by k to produce CP′. The image is then located relative to C.
Coordinates such as (5, 3) ordinarily describe displacement from the origin. They do not describe displacement from C = (2, 1). Subtracting the centre’s coordinates makes that change of reference explicit. Once the displacement has been scaled, add the centre back.
With C = (2, 1) and P = (5, 3), subtract coordinate by coordinate: P − C = (3, 2). Multiply this displacement by two to get (6, 4). Add C to obtain (2 + 6, 1 + 4) = (8, 5).
The original displacement has length √13, while the image displacement has length 2√13. Their directions agree because the factor is positive. This checks the geometry behind the coordinate calculation. The centre itself would have displacement zero and therefore remain at (2, 1).
Chapter 3 of 8 · Enlarge every vertex
3. Why does multiplying coordinates use the origin?
The shortcut P′ = kP is the special case C = (0, 0). In the general formula C + k(P − C), a zero centre makes both centre terms vanish. If the stated centre is nonzero, those terms normally matter and cannot simply be omitted.
For the main example, doubling coordinates gives (10, 6). Its displacement from C is (8, 5), which is not twice the original displacement (3, 2). It therefore fails the required centre-based test. A point can look farther away and still be the wrong image.
Apply the same centre and factor to every vertex. With centre (2, 1), factor two and vertices P = (5, 3), Q = (2, 4), R = (4, 1), the images are P′ = (8, 5), Q′ = (2, 7) and R′ = (6, 1).
Join the matched image vertices in the same order. Corresponding side lengths double because differences between image vertices are twice the original differences. Do not move the centre after calculating the first vertex. Every point belongs to one transformation with one stated reference and factor.
| Original vertex | Displacement from (2, 1) | Doubled displacement | Image vertex |
|---|---|---|---|
| P = (5, 3) | (3, 2) | (6, 4) | (8, 5) |
| Q = (2, 4) | (0, 3) | (0, 6) | (2, 7) |
| R = (4, 1) | (2, 0) | (4, 0) | (6, 1) |
Chapter 5 of 8 · Check factors and inverses
5. What do fractional and negative factors change?
For the same C and P, k = 1/2 gives C + (1/2)(3, 2) = (3.5, 2). The image lies halfway from the centre to P and the shape’s lengths are reduced by a factor of one half. Enlargement is the transformation’s name even when the image is smaller.
For k = −1, P′ = C − (3, 2) = (−1, −1). The image lies on the opposite side of the centre, at the same distance. A negative factor does not produce negative side lengths; lengths scale by |k| while directed displacement also reverses.
Chapter 6 of 8 · Check factors and inverses
6. How can the result be checked or reversed?
Check that P − C and P′ − C are scalar multiples with the stated factor. For positive k, the points lie on the same ray from the centre; for negative k, on opposite rays. Also check that the centre remains fixed and corresponding lengths scale by |k|.
For nonzero k, the inverse enlargement uses the same centre and factor 1/k. Applying factor one half about (2, 1) to image (8, 5) returns (5, 3). A zero factor collapses all points to the centre and cannot be reversed to recover distinct original points, so the reciprocal method does not apply to it.
Use centre C = (1, −2) and original point P = (4, 2). Question 1: find the image for factor two. Question 2: find it for factor one half. Question 3: find it for factor negative one. Question 4: where does the centre itself move under any of these transformations?
Answers: P − C = (3, 4). Therefore 1) (7, 6). 2) (2.5, 0). 3) (−2, −6). 4) The centre stays at (1, −2). Ask for subtract, scale and add as three visible steps, then check the displacement ratio.
Bring the coordinates, the stated centre and the child’s proposed image. A precise note is ‘uses origin-centred coordinate multiplication for a different centre’. In a small-group 3-pax lesson, students can enlarge the same point from two centres and explain why the image changes despite an unchanged factor.
For parents considering Secondary 4 Mathematics tuition in Punggol, the useful prompt is ‘From which point are these distances being measured?’ Once that reference is secure, change the factor or centre to check transfer. The goal is to make coordinate operations express the geometry that the question actually specifies.
