Locus Theorem: (two points)

magentarita · #1 $Old$ 11-15-2008, 11:18 PM

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.

Soroban · #2 $Old$ 11-16-2008, 02:24 PM

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

magentarita · #3 $Old$ 11-16-2008, 11:39 PM

Quote:

Originally Posted by Soroban $View Post$

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!