perpendicular bisector

Let $\overline{AB}$ be a line segment in a plane (we are assuming the Euclidean plane). A bisector of $\overline{AB}$ is any line that passes through the midpoint of $\overline{AB}$. A perpendicular bisector of $\overline{AB}$ is a bisector that is perpendicular to $\overline{AB}$.

It is an easy exercise to show that a line $\mathrm{\ell }$ is a perpendicular bisector of $\overline{AB}$ iff every point lying on $\mathrm{\ell }$ is equidistant from $A$ and $B$. From this, one concludes that the perpendicular bisector of a line segment is always unique.

A basic way to construct the perpendicular bisector $\mathrm{\ell }$ given a line segment $\overline{AB}$ using the standard ruler and compass construction is as follows:

1. use a compass to draw the circle $C_1$ centered at point $A$ with radius the length of $\overline{AB}$, by fixing one end of the compass at $A$ and the movable end at $B$,

2. similarly, draw the circle $C_2$ centered at $B$ with the same radius as above, with the compass fixed at $B$ and movable at $A$,

3. $C_1$ and $C_2$ intersect at two points, say $P,Q$ (why?)

4. with a ruler, draw the line $\overleftrightarrow{PQ}=\mathrm{\ell }$,

5. then $\mathrm{\ell }$ is the perpendicular bisector of $\overline{AB}$.

Figure 1: The construction of a perpendicular bisector

(Note: we assume that there is always an ample supply of compasses and rulers of varying sizes, so that given any positive real number $r$, we can find a compass that opens wider than $r$ and a ruler that is longer than $r$).

One of the most common use of perpendicular bisectors is to find the center of a circle constructed from three points in a Euclidean plane:

Given three non collinear points $X,Y,Z$ in a Euclidean plane, let $C$ be the unique circle determined by $X,Y,Z$. Then the center of $C$ is located at the intersection of the perpendicular bisectors of $\overline{XY}$ and $\overline{YZ}$.