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### Point Equidistant from 3 Other Points

```
Date: 04/11/99 at 09:40:10
From: Anonymous
Subject: Algebra

How do you find a point that is equidistant from three other points
using algebra?
```

```
Date: 04/11/99 at 12:56:03
From: Doctor Maryanthe
Subject: Re: Algebra

Hi, and thanks for writing Dr. Math!

Before solving an algebra problem, it sometimes helps to get a
geometric picture of what's happening. Geometry says that three points
determine a circle - in other words, given three points that are not
all on the same line, there is exactly one circle which passes through
all 3. Finding the point equidistant from the 3 points is the same
thing as finding the center of the circle that passes through all of
them (since all points on a circle are equidistant from the center).

Let's call our points A, B and C. Draw the lines AB, AC and BC to
form a triangle. Now construct the perpendicular bisectors of any two
of the lines, and their intersection will be the center of this
circle.

For a picture of this, see Clark Kimberling's Triangle Centers page:

http://cedar.evansville.edu/~ck6/tcenters/class/ccenter.html

(Click on "triangle centers" for a lot of other interesting stuff

Note: the "perpendicular bisector" of a line segment X is the line
that is perpendicular to X and passes through the midpoint of X.

Algebraically, then, what we want to do is:

1) Given the coordinates of A and B, find the equation for the line
segment which goes from A to B
2) Find the midpoint of that line segment
3) Find the equation of the line which is perpendicular to the line
in 1) and passes through the point we found in 2)
4) Repeat steps 1-3 with B and C instead of A and B
5) Find the point at which the two perpendicular bisectors cross.
This is the answer.

If you have any questions about how to do these steps, write back. I
hope this helps.

- Doctor Maryanthe, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Euclidean/Plane Geometry
High School Geometry

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