Isometries
Senior 2 · Mathematics
Isometries
Lesson 2 of 5
20 min
On a calm morning, the hills around Lake Kivu appear upside down in the water. A leaf, a butterfly, the front of many houses and much imigongo art are symmetrical: one half is the mirror image of the other. A reflection describes that mirror image exactly, point by point.
A reflection in a line m (the mirror line, or axis of reflection) maps each point A to the point A′ such that m is the perpendicular bisector of AA′. In other words, AA′ crosses m at right angles, and A and A′ are the same distance from m, on opposite sides.
Points on the mirror line do not move: they are fixed points.
To construct A′: slide a set square along the mirror line until one edge passes through A, draw the perpendicular from A to m, measure the distance, and mark A′ the same distance on the other side.
| Mirror line | Rule |
|---|---|
| the x-axis | (x, y) → (x, −y) |
| the y-axis | (x, y) → (−x, y) |
| the line y = x | (x, y) → (y, x) |
| the line y = −x | (x, y) → (−y, −x) |
| the vertical line x = a | (x, y) → (2a − x, y) |
| the horizontal line y = b | (x, y) → (x, 2b − y) |
Worked example 1
Reflect triangle A(1, 1), B(4, 1), C(1, 3) in the y-axis.
Change the sign of x: A′(−1, 1), B′(−4, 1), C′(−1, 3).
Figure 2. Reflection in the y-axis: each dashed segment is cut at right angles and in half by the mirror line. Diagram: Ganzaa.
Worked example 2
Reflect P(5, 2) in the line x = 3, then separately in the line y = x.
In x = 3: P is 2 units to the right of the line, so P′ is 2 units to the left: (2 × 3 − 5, 2) = (1, 2).
In y = x: swap the coordinates: (2, 5).
Worked example 3
A reflection maps A(2, 4) to A′(2, −2). Find the mirror line.
AA′ is vertical, so the mirror line is horizontal and passes through the midpoint (2, 1). The mirror line is y = 1.
Tip
Perpendicular, and the same distance. Every reflection question comes back to these two facts about the mirror line. For the line y = x, just swap the coordinates; for y = −x, swap them and change both signs.
Reflect (3, −7) in the x-axis.
Solution: (3, 7).
Reflect (3, −7) in the line y = −x.
Solution: swap and change both signs: (7, −3).
Reflect (−2, 5) in the line y = 1.
Solution: (−2, 2 × 1 − 5) = ( − 2, −3).
A reflection maps A(0, 6) to A′(4, 2). Find the mirror line.
Solution: the midpoint of AA′ is (2, 4). AA′ = 4−4 has gradient −1, so the mirror line, perpendicular to it, has gradient 1: y = x + 2. Check: the point (2, 4) lies on it, because 2 + 2 = 4.
The vertices A, B, C of a triangle go round clockwise. Which way do A′, B′, C′ go after a reflection?
Solution: anticlockwise, because a reflection reverses orientation.
Remember
Sources and credits
The Rwandan examples, practice items and exercises are written by Ganzaa.
