Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.2 Linear and Quadratic Functions - Exercises - Page 18: 21

Answer

$y=-2x+9$

Work Step by Step

$Q= (x_{1},y_{1})=(\frac{1+5}{2},\frac{2+4}{2})=(3,3)$ Slope of the segment= $\frac{4-2}{5-1}=\frac{1}{2}$ Slope of perpendicular bisector m= $\frac{-1}{slope\,of\,the\,segment}=\frac{-1}{\frac{1}{2}}=-2$ Equation of a line in point-slope form is $y-y_{1}=m(x-x_{1})$ Plugging in the values, we have $y-3=-2(x-3)$ or $y=-2x+9$
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