perpendicular bisectorPerpendicularBisector2006-12-22 12:14:382009-02-09 18:14:10Definitionbisector\usepackage{amssymb,amscd}
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Let $\overline{AB}$ be a line segment in a plane (we are assuming the Euclidean plane). A \emph{bisector} of $\overline{AB}$ is any line that passes through the midpoint of $\overline{AB}$. A \emph{perpendicular bisector} of $\overline{AB}$ is a bisector that is perpendicular to $\overline{AB}$.
It is an easy exercise to show that a line $\ell$ is a perpendicular bisector of $\overline{AB}$ iff every point lying on $\ell$ is equidistant from $A$ and $B$. From this, one concludes that \emph{the} perpendicular bisector of a line segment is always unique.
A basic way to construct the perpendicular bisector $\ell$ given a line segment $\overline{AB}$ using the standard ruler and compass construction is as follows:
\begin{enumerate}
\item 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 $B$,
\item similarly, draw the circle $C_2$ centered at $B$ with the same radius as above, with the compass fixed at $B$ and movable at $A$,
\item $C_1$ and $C_2$ intersect at two points, say $P,Q$ (why?)
\item with a ruler, draw the line $\overleftrightarrow{PQ}=\ell$,
\item then $\ell$ is the perpendicular bisector of $\overline{AB}$.
\end{enumerate}
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\caption{The construction of a perpendicular bisector}
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(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:
\begin{quote}
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 $\overline{YZ}$.
\end{quote}