## Calculus 10th Edition

Published by Brooks Cole

# Chapter P - P.2 - Linear Models and Rates of Change - Exercises - Page 17: 69

#### Answer

$(0,\frac{-a^2+b^2+c^2}{2c})$

#### Work Step by Step

To find the intersection, we must find the equation of each perpendicular bisector. First find the midpoint and slope of each side. Then using that information we can find each perpendicular bisector's slope and equation. 1. Find the perpendicular bisector of the base of the triangle: $midpoint=(\frac{-a+a}{2},\frac{0+0}{2})=(0,0)$ $slope=\frac{0-0}{a+a}=0$ which is perpendicular to a vertical line Equation of the perpendicular bisector: $x=0$ 2. Find the perpendicular bisector of the left side: $midpoint=(\frac{-a+b}{2},\frac{c+0}{2})=(\frac{-a+b}{2},\frac{c}{2})$ $slope=\frac{c-0}{b+a}=\frac{c}{b+a}$ which is perpendicular to $-\frac{b+a}{c}$ Equation of the perpendicular bisector in point-slope form: $y-\frac{c}{2}=-\frac{b+a}{c} (x-\frac{-a+b}{2})$ 3. Find the perpendicular bisector of the right side: $midpoint = (\frac{a+b}{2},\frac{c+0}{2})= (\frac{a+b}{2},\frac{c}{2})$ $slope=\frac{c-0}{b-a}=\frac{c}{b-a}$ which is perpendicular to $\frac{a-b}{c}$ Equation of the perpendicular bisector in point-slope form: $y-\frac{c}{2}=\frac{a-b}{c} (x-\frac{a+b}{2})$ Now we can find the intersection by combining any two of these equations. I will use the equation found in step 1 and 3: $y-\frac{c}{2}=\frac{a-b}{c} (x-\frac{a+b}{2})$ $y-\frac{c}{2}=\frac{a-b}{c} (0-\frac{a+b}{2})$ [substitute x = 0] $y-\frac{c}{2}=\frac{b-a}{c} (\frac{b+a}{2})$ $y-\frac{c}{2}=\frac{b^2-a^2}{2c}$ [difference of squares] $y-\frac{c^2}{2c}=\frac{b^2-a^2}{2c}$ $y=\frac{b^2-a^2}{2c}+\frac{c^2}{2c}$ $y=\frac{-a^2+b^2+c^2}{2c}$ The point of intersection is $(0,\frac{-a^2+b^2+c^2}{2c})$

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