# Thread: Locus Theorem: (two points)

1. ## 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.

2. Hello, magentarita!

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

3. ## great............

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!