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 about triangles.) 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/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum