GeometryRigid MotionsG-CO.A.4

Precise Definitions of Rigid Motions

Define rotations, reflections and translations with angles, circles, perpendiculars and parallels.

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In Transformations as Functions you used coordinate rules. To prove things about rigid motions, we need definitions that do not depend on a grid, built from tools you already have: circles, perpendicular bisectors and parallel lines from Constructing Perpendicular and Parallel Lines.

  • Translation by a vector $\vec{v}$: every point moves the same distance in the same direction, so each segment $\overline{PP'}$ is parallel to $\vec{v}$ and has the same length.
  • Reflection over a line $\ell$: a point on $\ell$ stays put; any other point $P$ goes to the point $P'$ for which $\ell$ is the perpendicular bisector of $\overline{PP'}$.
  • Rotation about a center $C$ by an angle $\theta$: $C$ stays put; any other point $P$ goes to the point $P'$ with $CP' = CP$ (same circle around $C$) and $m\angle PCP' = \theta$, counterclockwise when $\theta$ is positive.

Why it works

These definitions use only ideas we can construct and reason about. That is what lets us prove that rigid motions keep distances, and later that they explain congruence. For example, a reflection keeps $PQ$ because the two halves of the picture on either side of $\ell$ are mirror images made of congruent right triangles.
Pick a reflection line and watch: the line is always the perpendicular bisector of each segment joining a point to its image.

Worked example: Reflecting over a vertical line

Reflect $P(5, 1)$ over the line $x = 2$.
  1. $P$ is $3$ units right of the line, so $P'$ is $3$ units left of it: $x = 2 - 3 = -1$.
  2. The line is vertical, so $\overline{PP'}$ is horizontal and $y$ does not change.
  3. $P' = (-1, 1)$. Check: the midpoint of $\overline{PP'}$ is $(2, 1)$, on the line, and $\overline{PP'}$ is perpendicular to it.

Answer: $P'(-1, 1)$

Worked example: Rotating about a point that is not the origin

Rotate $P(3, 1)$ by $90^\circ$ counterclockwise about $C(1, 2)$.
  1. Relative to $C$, $P$ is $\langle 2, -1 \rangle$.
  2. A $90^\circ$ counterclockwise turn sends $\langle a, b \rangle$ to $\langle -b, a \rangle$: $\langle 1, 2 \rangle$.
  3. Add back $C$: $P' = (1 + 1, 2 + 2) = (2, 4)$ (shown as $Q$). Check: $CP = CP' = \sqrt{5}$ and the angle at $C$ is a right angle.

Answer: $P'(2, 4)$

Worked example: Finding the mirror

A reflection sends $P(1, 4)$ to $P'(5, 4)$. What is the line of reflection?
  1. The line is the perpendicular bisector of $\overline{PP'}$.
  2. Midpoint: $(3, 4)$. $\overline{PP'}$ is horizontal, so the bisector is vertical.

Answer: The line $x = 3$.

Key idea

Translation: every point moves by the same vector. Reflection: the mirror line is the perpendicular bisector of each $\overline{PP'}$. Rotation: each point stays on its circle around the center and turns through the same angle.

Common mistake

Rotating about the origin when the center is somewhere else. The rule $(x, y) \to (-y, x)$ only works for the origin. For another center, first find each point's position relative to the center, rotate that, then add the center back. Check: the center-to-point distance must not change.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.

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Sources and further reading

  • OpenStax Contemporary Mathematics, 10.5 Tessellations
  • Common Core State Standards for Mathematics, High School Geometry, G-CO.A.4
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