Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.10 - Lines - 1.10 Exercises - Page 115: 83

Answer

$$y=x-3$$

Work Step by Step

First, we have to find slope of the line connecting $A(1,4)$ and $B(7,-2)$ $m_{AB}=\frac{-2-4}{7-1}=\frac{-6}{6}=-1$ To find the equation, we have to first find slope of the perpendicular bisector of this line (Let the bisector be $b$). According to one of the slope rules, product of perpendicular line slopes is $-1$: $m_b \times m_{AB}=-1$ $m_b\times -1 = -1$ $m_b=1$ To write the equation, we also need any point, located on this line. Since we know, that it is a bisector, then one of its point will be in the middle of $A$ and $B$ (Let the point be $C$): $C(x, y)=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})=(\frac{1+7}{2}, \frac{4-2}{2})=(4, 1)$ We have $C(4,1)$ $m=\frac{y-y_0}{x-x_0}$ $y-y_0=m(x-x_0)$ $y-1=1(x-4)$ $$y=x-3$$
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