| Version 9 |
Version 8 |
| Let $\overline{AB}$ be a line segment in a plane (we are assuming the Euclidean plane). The \emph{perpendicular bisector} of $\overline{AB}$ is a line $\ell$ with the following properties: |
Let $\overline{AB}$ be a line segment in a plane (we are assuming the Euclidean plane). The \emph{perpendicular bisector} of $\overline{AB}$ is a line $\ell$ with the following properties: |
| \begin{itemize} |
\begin{itemize} |
| \item $\ell$ is perpendicular to $\overline{AB}$, and |
\item $\ell$ is perpendicular to $\overline{AB}$, and |
| \item $\ell$ passes through the midpoint of $\overline{AB}$ |
\item $\ell$ passes through the midpoint of $\overline{AB}$ |
| \end{itemize} |
\end{itemize} |
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| It is an easy exercise to show that $\ell$ is the perpendicular bisector of $\overline{AB}$ iff every point lying on $\ell$ is equidistant from $A$ and $B$. |
It is an easy exercise to show that $\ell$ is the perpendicular bisector of $\overline{AB}$ iff every point lying on $\ell$ is equidistant from $A$ and $B$. |
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| 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: |
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: |
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| \begin{enumerate} |
\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 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 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 $C_1$ and $C_2$ intersect at two points, say $P,Q$ (why?) |
| \item with a ruler, draw the line $\overleftrightarrow{PQ}=\ell$, |
\item with a ruler, draw the line $\overleftrightarrow{PQ}=\ell$, |
| \item then $\ell$ is the perpendicular bisector of $\overline{AB}$. |
\item then $\ell$ is the perpendicular bisector of $\overline{AB}$. |
| \end{enumerate} |
\end{enumerate} |
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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 $r$). |
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| \begin{center} |
\begin{center} |
| \begin{figure}[!htb] |
\begin{figure}[!htb] |
| \begin{center} |
\begin{center} |
| \includegraphics{construct.1.eps} |
\includegraphics{construct.1.eps} |
| \caption{The construction of a perpendicular bisector} |
\caption{The construction of a perpendicular bisector} |
| \end{center} |
\end{center} |
| \end{figure} |
\end{figure} |
| \end{center} |
\end{center} |
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| 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: |
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} |
\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 $\overline{YZ}$. |
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} |
\end{quote} |
