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Math Help - Locus Theorem: (two points)

  1. #1
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    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.


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  2. #2
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    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
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  3. #3
    MHF Contributor
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    great............

    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!
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