# Locus Theorem: (two points)

• Nov 15th 2008, 09:18 PM
magentarita
Locus Theorem: (two points)
The locus of points equidistant from two points, P and Q, is the perpendicular bisector of the line segment determined by the two points.

I need the above theorem explained an easier way.

• Nov 16th 2008, 12:24 PM
Soroban
Hello, magentarita!

Quote:

The locus of points equidistant from two points, $P$ and $Q,$
is the perpendicular bisector of the line segment determined by the two points.

We have two points, $P$ and $Q$, and the line segment joining them.
Code:

```     P *-----------* Q```

Find a point $A$ equidistant from $P$ and $Q.$
. . That is: . $AP \:=\:AQ$
Code:

```            A             o           * *           *  *         *    *         *      *       *        *     P *-----------* Q```

Find another point $B$ equidistant from $P$ and $Q.$
Code:

```            B             o           *  *         *      *     P *-----------* Q```

Find another point $C$ equidistant from $P$ and $Q.$
Code:

```     P *-----------* Q         *    *             o             C```

If we find all the points equidistant from $P$ and $Q$ (zillions of them),
. . they form the perpendicular bisector of segment $PQ.$
Code:

```             o             o             o             o             o             o     P *-----o-----* Q             o             o             o```
• Nov 16th 2008, 09:39 PM
magentarita
great............
Quote:

Originally Posted by Soroban
Hello, magentarita!

We have two points, $P$ and $Q$, and the line segment joining them.
Code:

```       P *-----------* Q```
Find a point $A$ equidistant from $P$ and $Q.$
. . That is: . $AP \:=\:AQ$
Code:

```            A             o           * *           *  *         *    *         *      *       *        *     P *-----------* Q```
Find another point $B$ equidistant from $P$ and $Q.$
Code:

```            B             o           *  *         *      *     P *-----------* Q```
Find another point $C$ equidistant from $P$ and $Q.$
Code:

```     P *-----------* Q         *    *             o             C```
If we find all the points equidistant from $P$ and $Q$ (zillions of them),
. . they form the perpendicular bisector of segment $PQ.$
Code:

```             o             o             o             o             o             o     P *-----o-----* Q             o             o             o```

Wonderfully done!