Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 10 - Section 10.5 - Equations of Lines - Exercises - Page 472: 45

Answer

For a triangle, the altitudes are always concurrent.

Work Step by Step

To prove that the altitudes 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 (b,c), and C the point (a,0). We find the equation of a line perpendicular to BC: $m_{BC}= \frac{c-0}{b-c} = \frac{c}{b-c}$ The altitude is perpendicular to this line an goes through (0,0), so we find that its equation is: $y =\frac{a-b}{c} x$ The slope of the other altitude is the inverse reciprocal of: $m_{AC} = \frac{c-0}{b-0}=\frac{c}{b}$ Thus: $y=\frac{-b}{c}(x-a)$ We now see where these lines intersect: $\frac{a-b}{c} x =\frac{-b}{c}(x-a) \\\frac{a-b}{-b} x = x-a \\ \frac{-a+b}{b} x - x = -a \\ \frac{-a+b-b}{b} x = -a \\ x = b$ Since the other altitude is the vertical line $x=b$, it follows that the altitudes of a triangle are concurrent.
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