A two-step transformation is a short journey with a stop in the middle. Losing that intermediate image is the easiest way to apply the right operations in the wrong order. The opening cases follow reflections and rotations; the later extension adds enlargements and matrix composition. Every vertex keeps its name, and every stage gets its own coordinates.
Work on equal-scale axes. Write A′ for the image of A after the first operation and A″ for its image after the second. Join corresponding vertices in the same order, A to B to C to A. Do not rename the leftmost point A merely because its position changed.
Write the instruction before moving a point
A reflection needs a named mirror line. A rotation needs a centre, an angle and a direction. “Rotate through 90°” is incomplete without the centre and direction. The following rules apply to the stated axes or origin; they cannot be transferred unchanged to a different centre.
| Specified operation | Point rule |
|---|---|
| Reflect in the x-axis | (x, y) → (x, −y) |
| Reflect in the y-axis | (x, y) → (−x, y) |
| Rotate 90° anticlockwise about (0, 0) | (x, y) → (−y, x) |
| Rotate 90° clockwise about (0, 0) | (x, y) → (y, −x) |
| Rotate 180° about (0, 0) | (x, y) → (−x, −y) |
For a reflection, the mirror bisects the segment joining a point to its image at right angles. For a rotation, the point and image stay the same distance from the centre. These geometric facts provide checks independent of memorising a coordinate rule.
Case 1: reflection followed by a quarter-turn
Start with the scalene triangle A(1, 2), B(4, 2), C(1, 4). First reflect it in the x-axis. Then rotate the image through 90° anticlockwise about O(0, 0). Track A first: (1, 2) → (1, −2) → (2, 1). The rotation acts on the reflected point, not on A again.
| Vertex | Original | After x-axis reflection | After 90° anticlockwise rotation |
|---|---|---|---|
| A | (1, 2) | (1, −2) | (2, 1) |
| B | (4, 2) | (4, −2) | (2, 4) |
| C | (1, 4) | (1, −4) | (4, 1) |
The original perpendicular sides AB and AC have lengths 3 and 2. In the final image, A″B″ is vertical with length 3, and A″C″ is horizontal with length 2. The third side is √13 at every stage. The area is therefore always ½ × 3 × 2 = 3 square units.
Reverse the operations and compare
Now rotate the original triangle 90° anticlockwise about the origin first, then reflect in the x-axis. The original and intermediate columns must be rebuilt; reusing the earlier intermediate image would repeat the same order.
| Vertex | After rotation first | After reflection second | Earlier final image |
|---|---|---|---|
| A | (−2, 1) | (−2, −1) | (2, 1) |
| B | (−2, 4) | (−2, −4) | (2, 4) |
| C | (−4, 1) | (−4, −1) | (4, 1) |
The two answers are different. In general, the first order sends (x, y) to (y, x), while the reversed order sends it to (−y, −x). This proves an order difference for the rules, rather than relying on the look of one triangle. A point such as the origin can still have the same image under both.
Case 2: two reflections that do commute
Reflect the original triangle in the x-axis and then in the y-axis. The first step changes y to −y; the second changes x to −x. Reversing those two steps gives the same final rule, (x, y) → (−x, −y).
| Vertex | Original | After x-axis reflection | After y-axis reflection |
|---|---|---|---|
| A | (1, 2) | (1, −2) | (−1, −2) |
| B | (4, 2) | (4, −2) | (−4, −2) |
| C | (1, 4) | (1, −4) | (−1, −4) |
The combined result is a 180° rotation about the origin. Two reflections reverse the orientation twice, restoring it. This case is a useful warning against the opposite overgeneralisation: although order often matters, it does not always matter.
Case 3: a rotation centre away from the origin
Rotate the same triangle 180° about K(2, 1), then reflect in the vertical line x = 1. A half-turn makes K the midpoint of each point-image segment. Thus the first rule is (x, y) → (4 − x, 2 − y). Reflection in x = 1 then changes an input horizontal coordinate u to 2 − u, while keeping its vertical coordinate.
| Vertex | Original | After half-turn about K | After reflection in x = 1 |
|---|---|---|---|
| A | (1, 2) | (3, 0) | (−1, 0) |
| B | (4, 2) | (0, 0) | (2, 0) |
| C | (1, 4) | (3, −2) | (−1, −2) |
For A, the midpoint of (1, 2) and (3, 0) is (2, 1), as required. Its intermediate and final x-coordinates, 3 and −1, average to 1. Together the two rules give (x, y) → (x − 2, 2 − y), which agrees with every row.
Diagnose a wrong-order answer
A learner gives A″(−2, −1) for Case 1 and says the triangle still has area 3, so the answer must be right. The area check passes because either sequence preserves area. However, (−2, −1) belongs to the reversed sequence. Ask for A′: the first instructed step must produce (1, −2). That single intermediate point exposes the error.
Three solved practice tasks
- Reflect P(−3, 2) in the y-axis, then rotate 90° clockwise about the origin. Reflection gives P′(3, 2). The clockwise rule sends this to P″(2, −3). Its distance from the origin remains √13.
- Rotate Q(5, −1) by 180° about (2, 1), then reflect in x = 1. Q′ = (4 − 5, 2 − (−1)) = (−1, 3). Then Q″ = (2 − (−1), 3) = (3, 3). The midpoint check for the first step gives (2, 1).
- Reflect R(2, −5) in the y-axis and then in the x-axis. Compare the reverse order. The first sequence is (2, −5) → (−2, −5) → (−2, 5). The reverse sequence is (2, −5) → (2, 5) → (−2, 5). They agree, as the general rule predicts.
Finish a full-figure answer by checking all vertices, not just the first. For the reflection and rotation cases above, preserve side lengths and angles. For enlargement, check the stated length and area scale factors. Expect one orientation reversal for each reflection, and write the final transformation order beside the finished figure.
Add enlargement and matrix composition
Reflections and rotations preserve lengths. An enlargement needs a centre K and a scale factor k. Its point rule is P′ = K + k(P − K). Directed distances from K are multiplied by k; lengths by |k| and areas by k². For k < 0, the image is on the opposite ray from K. The cases below use nonzero k.
Worked enlargement: move relative to the stated centre
Enlarge triangle A(1,2), B(4,2), C(1,4) by scale factor 2 about K(1,1), then reflect in the x-axis. For A, the offset from K is (0,1); doubling gives (0,2), so A′ = (1,3). The final reflection gives A″ = (1,−3). The rule for the enlargement is (x,y) → (2x − 1, 2y − 1).
| Vertex | Original | After enlargement | After reflection |
|---|---|---|---|
| A | (1,2) | (1,3) | (1,−3) |
| B | (4,2) | (7,3) | (7,−3) |
| C | (1,4) | (1,7) | (1,−7) |
The original perpendicular sides have lengths 3 and 2, so area = 3 square units. After enlargement they are 6 and 4, with area 12 square units. Reflection preserves that area. Reversing these two operations sends A through (1,−2) to (1,−5), so it does not give the same final image. The non-origin centre is essential to this check.
Worked matrix composition: the rightmost matrix acts first
Use coordinate columns [x; y]. A semicolon separates rows: [a, b; c, d][x; y] = [ax + by; cx + dy]. Reflection in the x-axis has matrix F = [1, 0; 0, −1], a 90° anticlockwise rotation about the origin has R = [0, −1; 1, 0], and enlargement by 2 about the origin has E = [2, 0; 0, 2].
Reflect, then rotate, then enlarge: P″′ = E(R(FP)) = (ERF)P. First RF = [0, 1; 1, 0]; then ERF = [0, 2; 2, 0]. For example, its top row is [2,0] multiplied by the columns of RF, giving [0,2]. Thus the full point rule is (x,y) → (2y,2x).
| Vertex | After F | After R | After E |
|---|---|---|---|
| A(1,2) | (1,−2) | (2,1) | (4,2) |
| B(4,2) | (4,−2) | (2,4) | (4,8) |
| C(1,4) | (1,−4) | (4,1) | (8,2) |
Applying the composite matrix directly gives exactly the last column. The final perpendicular sides are 6 and 4, so area is again 12. Reversing just F and R gives FR = [0, −1; −1, 0], so EFR sends A to (−4,−2), not (4,2). Although the origin-centred enlargement E commutes with F and R, that does not make F and R commute.
A two-by-two matrix acting on [x; y] always fixes (0,0). It cannot by itself represent enlargement about K(1,1) above, which sends the origin to (−1,−1). Use the centre-offset rule, or subtract K, multiply the offset by E, then add K back. Do not silently replace the centre by the origin.
Independent transformation tasks
- Enlarge P(3,1) by factor −2 about K(1,1), then rotate its image 90° clockwise about the origin. State the intermediate and final coordinates.
- For triangle U(1,0), V(3,0), W(1,2), first rotate 90° anticlockwise about the origin, then reflect in the y-axis using G = [−1, 0; 0, 1], then enlarge by 1/2 about the origin. Find the composite matrix, all image vertices and the final area.
Explained answers
- P − K = (2,0); −2(P − K) = (−4,0); add K to get P′ = (−3,1). The clockwise rule (x,y) → (y,−x) gives P″ = (1,3). The distance from K is doubled by the enlargement, and P′ is on the opposite ray.
- The composite is (1/2)I · G · R = [0, 1/2; 1/2, 0]. The images are U′ = (0,1/2), V′ = (0,3/2), W′ = (1,1/2). Original area = ½ × 2 × 2 = 2; final area = (1/2)² × 2 = 1/2 square unit. Directly, the final perpendicular sides are 1 and 1, giving the same area.
Frequently asked questions
What does “followed by” tell me?
It gives the order. Apply the first operation to the original figure, then apply the second operation to the intermediate image. Keep a separate coordinate column for each stage.
What information completely specifies a rotation?
State its centre, angle and direction. For a half-turn, clockwise and anticlockwise give the same image, but the centre still matters.
Can I reflect by changing both coordinate signs?
Changing both signs gives a half-turn about the origin. Reflection in the x-axis changes only the y-coordinate; reflection in the y-axis changes only the x-coordinate.
Must reversing the order change the answer?
No. The x-axis and y-axis reflections in Case 2 commute. Case 1 gives different final coordinates, so there is no general permission to reverse an arbitrary pair.
Why use equal scales on the coordinate axes?
Equal physical lengths must represent equal coordinate units. Unequal scales can make a correct right angle look wrong or make a rigidly moved triangle appear stretched.
What does orientation mean here?
It is the clockwise or anticlockwise order of the named vertices around the triangle. A reflection reverses that order; a rotation preserves it. It is different from where the triangle points.
Is matching area enough to prove the transformation is correct?
No. Many different positions and shapes can have the same area. Check the rule, all corresponding coordinates, the expected length and area scale factors, and the expected orientation change.
Further reading
For the underlying mathematics, see AMSI’s explanation of rigid motions and congruence. The mathematics topic-test collection is a separate resource page that currently shows a member login; specific files and their answers are not verified here.
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