Isometries
Senior 2 · Mathematics
Isometries
Lesson 3 of 5
22 min
The hands of a clock, the wheel of a bicycle, a door on its hinges and the blades of a ceiling fan all turn about a fixed point. Many imigongo and basket patterns are made by turning one motif a quarter-turn at a time. A rotation describes such a turn precisely.
A rotation is fixed by three things:
The image A′ of a point A satisfies OA′ = OA and angle AOA′ = θ, measured in the given direction.
To construct A′: join O to A; place a protractor on OA with its centre at O and mark the angle θ in the right direction; draw a ray from O through the mark; open your compasses to the length OA and mark A′ on the ray. Tracing paper, pinned at O and turned, is a good way to check.
| Rotation about O | Rule |
|---|---|
| +90° (a quarter-turn anticlockwise) | (x, y) → (−y, x) |
| 180° (a half-turn, either way) | (x, y) → (−x, −y) |
| −90° (a quarter-turn clockwise, the same as +270°) | (x, y) → (y, −x) |
Worked example 1
Rotate triangle A(2, 1), B(5, 1), C(2, 3) through +90° about the origin.
Using (x, y) → (−y, x): A′(−1, 2), B′(−1, 5), C′(−3, 2).
Check one distance: OA = √(2² + 1²) = √5 and OA′ = √((−1)² + 2²) = √5.
Figure 3. A rotation of 90° anticlockwise about O. OA and OA′ are equal and the angle AOA′ is 90°. Diagram: Ganzaa.
Worked example 2
Rotate P(3, −2) through 90° clockwise about the origin.
Using (x, y) → (y, −x): P′ = (−2, −3).
Worked example 3
(a centre that is not the origin). Rotate P(5, 2) through +90° about C(1, 1).
Step 1, the vector from C to P: 5 − 12 − 1 = 41.
Step 2, rotate that vector: 41 → −14.
Step 3, add it back to C: P′ = (1 − 1, 1 + 4) = (0, 5).
Check: CP = √(16 + 1) = √17 and CP′ = √(1 + 16) = √17.
Tip
For +90° about O: swap the coordinates, then change the sign of the new first one. (2, 1) → (1, 2) → (−1, 2). For −90°, swap and change the sign of the new second one. About any other centre, always work with the vector from the centre, rotate it, and add it back.
Rotate (6, 1) through +90° about the origin.
Solution: (−1, 6).
Rotate (6, 1) through 180° about the origin.
Solution: (−6, −1).
Rotate (−2, 5) through 90° clockwise about the origin.
Solution: (x, y) → (y, −x) gives (5, 2).
Rotate (4, 4) through +90° about C(2, 1).
Solution: the vector from C is 23; rotated it becomes −32; so the image is (2 − 3, 1 + 2) = ( − 1, 3).
Through what angle does the minute hand of a clock turn in 20 minutes? Give the sign.
Solution: it turns 6° per minute, so 20 × 6° = 120°. Clocks turn clockwise, so the rotation is −120°.
Remember
Sources and credits
The Rwandan examples, practice items and exercises are written by Ganzaa.
