Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 3 - Section 3.5 - Equations of Lines - Exercise Set - Page 175: 92

Answer

$x-y=-5$

Work Step by Step

Using $\left( \dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2} \right)$ or the Midpoint Formula, then the midpoint of the line segment joining the points $ (-6,-3) \text{ and } (-8,-1) $ is \begin{array}{l}\require{cancel} \left( \dfrac{-6+(-8)}{2},\dfrac{-3+(-1)}{2} \right) \\\\ \left( \dfrac{-6-8}{2},\dfrac{-3-1}{2} \right) \\\\ \left( \dfrac{-14}{2},\dfrac{-4}{2} \right) \\\\ \left( -7,-2 \right) .\end{array} Using $m=\dfrac{y_1-y_2}{x_1-x_2}$ or the Slope Formula, then \begin{array}{l}\require{cancel} m=\dfrac{-3-(-1)}{-6-(-8)} \\\\ m=\dfrac{-3+1}{-6+8} \\\\ m=\dfrac{-2}{2} \\\\ m=-1 .\end{array} Taking the negative reciprocal of $m$, then the slope of the perpendicular bisector is $m_p= 1 .$ Using $ (-7,-2) $ and $m_p= 1 ,$ the equation of the perpendicular bisector is \begin{array}{l}\require{cancel} y-(-2)=1(x-(-7)) \\\\ y+2=1(x+7) \\\\ y+2=x+7 \\\\ -x+y=7-2 \\\\ -x+y=5 \\\\ x-y=-5 .\end{array}
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