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: 94

Answer

$x-3y=-9$

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 $ (5,8) \text{ and } (7,2) $ is \begin{array}{l}\require{cancel} \left( \dfrac{5+7}{2},\dfrac{8+2}{2} \right) \\\\ \left( \dfrac{12}{2},\dfrac{10}{2} \right) \\\\ \left( 6,5 \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{8-2}{5-7} \\\\ m=\dfrac{6}{-2} \\\\ m=-3 .\end{array} Taking the negative reciprocal of $m$, then the slope of the perpendicular bisector is $m_p= \dfrac{1}{3} .$ Using $ (6,5) $ and $m_p= \dfrac{1}{3} ,$ the equation of the perpendicular bisector is \begin{array}{l}\require{cancel} y-5=\dfrac{1}{3}(x-6) \\\\ 3(y-5)=1(x-6) \\\\ 3y-15=x-6 \\\\ -x+3y=-6+15 \\\\ -x+3y=9 \\\\ x-3y=-9 .\end{array}
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