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Date: 05/03/99 at 08:37:49
From: Jo
Subject: Locus

What do you have to do to get the perpendicular bisector of all the 
loci of a triangle? What does it mean when it asks you about 
equiddistance from certain points?


Date: 05/03/99 at 13:02:03
From: Doctor Peterson
Subject: Re: Locus

Hi, Jo. Thanks for writing.

Your first question doesn't make a lot of sense, because there is no 
"locus of a triangle." But I think I can see what you're asking about.

A "locus" is the set of all points that satisfy some rule or 
description, and a perpendicular bisector is one of the simplest 
examples of a locus. If I asked you "what is the locus of all points 
equidistant from two given points?", the answer would be "the 
perpendicular bisector of the segment determined by the two points." 
Here's what the question means:

    What is the locus of all points ...

        (What geometrical object consists of all points X ...)

    that are equidistant from A and B?

        (... for which the distances AX and BX are equal?)

That is, "equidistant from two points" means "the same distance from 
both points."

Here's what the answer means:

    The perpendicular bisector ...

        (the line through the midpoint of the segment, perpendicular 
         to the segment)

    of the segment determined by A and B.

        (you make the segment by connecting A and B with a straight 

You can prove (and probably your text did this) that any point on the 
perpendicular bisector of a segment AB is the same distance from both 
endpoints; you can see this by drawing in the segments and seeing 
that you have an isosceles triangle, so that AX = BX:

                      + X
                    / | \
                   /  |  \
                  /   |   \
                 /    |    \
                /     |     \
               /      |      \
              /       |       \
             /        |        \
            /         |         \
           /          |          \
          /           |           \
         /            |_           \
        /             | |           \
       A              |              B

This tells us that any point on the perpendicular bisector is 
equidistant from A and B, so it is part of the locus we are looking 

Conversely, you can show that if any point is equidistant from A and 
B, it must be on the perpendicular bisector, so there are no other 
points in the locus: the perpendicular bisector IS the entire locus 
we're looking for.

That's what a locus is: the set of ALL points that fit some 
description (in this case, being equidistant from A and B). In our 
case, the perpendicular bisector is "the locus, the whole locus, and 
nothing but the locus" of equidistant points.

There are many other examples of a locus. For example, the locus of 
points 1 inch from point A is the circle of radius 1 centered at A. 
Every point of the circle is 1 inch from A, and every point one inch 
from A is part of the circle.

I hope that helps a little to clarify the idea of a locus, and the 
meaning of equidistant. If you have problems you still have trouble 
with, maybe you can write back with a specific problem so I can see 
the wording of it.

- Doctor Peterson, The Math Forum   
Associated Topics:
High School Definitions
High School Euclidean/Plane Geometry
High School Geometry
High School Triangles and Other Polygons

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