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Find a rotation and translation that map one triangle to another

Test one rule against every vertex

A $9 0^{\circ}$ counterclockwise rotation about the origin sends $(x, y)$ to $(- y, x)$ . A translation two units right then adds $2$ to the first coordinate. A transformation is valid only if it maps every corresponding vertex, not just one.

Problem

Triangle $A B C$ has vertices $A (- 3, 4)$ , $B (- 3, 0)$ and $C (- 1, 3)$ . Its image has vertices $A^{'} (- 2, - 3)$ , $B^{'} (2, - 3)$ and $C^{'} (- 1, - 1)$ . Describe a sequence of transformations mapping $A B C$ to $A^{'} B^{'} C^{'}$ and give the coordinate rules.

Answer

Rotate

9 0^{\circ}

counterclockwise about the origin, then translate

2

units right. Combined rule:

(x, y) \mapsto (- y + 2, x)

All three original vertices map to their named image vertices.

Step-by-step solution

1. Try a quarter-turn

A $9 0^{\circ}$ counterclockwise rotation sends $(x, y)$ to $(- y, x)$ . It maps $A (- 3, 4)$ to $(- 4, - 3)$ , which is two units left of $A^{'} (- 2, - 3)$ .

2. Add a translation

Move every rotated point two units right: $(u, v) \mapsto (u + 2, v)$ . The combined rule is

(x, y) \mapsto (- y + 2, x) .

3. Check all vertices

$A (- 3, 4) \mapsto (- 2, - 3)$ , $B (- 3, 0) \mapsto (2, - 3)$ and $C (- 1, 3) \mapsto (- 1, - 1)$ . These match $A^{'}$ , $B^{'}$ and $C^{'}$ exactly.

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