College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.3 - Lines - 1.3 Exercises - Page 114: 83

Answer

$x-y-3=0$

Work Step by Step

First we determine the coordinates of the midpoint $M(x_M,y_M)$ of $AB$: $$\begin{align*} x_M&=\dfrac{x_A+x_B}{2}=\dfrac{1+7}{2}=4\\ y_M&=\dfrac{y_A+y_B}{2}=\dfrac{4+(-2)}{2}=1. \end{align*}$$ Therefore we have to determine the equation of the line perpendicular on $AB$ and passing through the point $M(4,1)$. We determine the slope $m_{AB}$ of the line passing through $A$ and $B$: $$m_{AB}=\dfrac{y_B-y_A}{x_B-x_A}=\dfrac{-2-4}{7-1}=-1.$$ Determine the slope $m$ of the line perpendicular to $AB$: $$\begin{align*} mm_{AB}&=-1\\ m\cdot (-1)&=-1\\ m&=1. \end{align*}$$ Therefore the equation of the perpendicular on $AB$ is: $$y=x+b.$$ In order to find $b$ use the coordinates of the point $M$ which belongs to this line: $$\begin{align*} 1&=4+b\\ b&=-3. \end{align*}$$ The equation of the perpendicular bisector is: $$\begin{align*} y=x-3\\ x-y-3&=0. \end{align*}$$
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