Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10: 9781285195698
ISBN 13: 978-1-28519-569-8

Chapter 10 - Section 10.5 - Equations of Lines - Exercises - Page 472: 46


The perpendicular bisectors are concurrent, for they all intersect at a common point for a triangle.

Work Step by Step

To prove that the perpendicular bisectors are concurrent, we must prove that they intersect at the same point. We first define coordinates, calling A the point (0,0), B the point (2b,2c), and C the point (2a,0). We find the equation of a line perpendicular to BC: $m_{BC}= \frac{2c-0}{2b-2a} = \frac{c}{b-a}$ The equation of the perpendicular bisector is: $y - c =\frac{b-c}{-c} (x -a-b)$ The slope of the bisector is the inverse reciprocal of: $m_{AC} = \frac{c-0}{b-0}=\frac{c}{b}$ Thus: $y - c=\frac{-b}{c}(x-b)$ We now see where these lines intersect: $\frac{-b}{c}(x-b) = \frac{b-c}{-c} (x -a-b) \\ x-b = \frac{b-c}{b}(x-a-b) \\ x-b =\frac{b-c}{b}x - \frac{b-c}{b}a -b+c\\ x -\frac{b-c}{b}x=- \frac{b-c}{b}a+c \\ \frac{c}{b} x = \frac{-ba+ca+cb}{b} \\ x = a$ a is the midpoint of the first line, which is where the vertical perpendicular bisector goes through, so we see that they all intersect at a common point.
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