Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 1 - Linear Functions - 1.1 Slopes and Equations of Lines - 1.1 Exercises - Page 13: 34

Answer

$\frac{1}{2}x+y=\frac{-1}{3}$

Work Step by Step

Perpendicular lines have slopes that are negative reciprocals of each other. Write the equation in slope-intercept form $y=2x-4$ The given line has a slope of $2$ This means that the slope of the line perpendicular to it is $\frac{-1}{2}$ Thus, the tentative equation of the line is $y=\frac{-1}{2}x+b$ The line is said to have a x-intercept of $\frac{-2}{3}$ so $b=\frac{-2}{3}$ Therefore the equation of the line perpendicular to the given line is $y=\frac{-1}{2}(x-\frac{-2}{3})$ Convert this to $ax+by=c$ form to have: $y=\frac{-1}{2}x-\frac{2}{6}$ $\frac{1}{2}x+y=\frac{-2}{6}$ $\frac{1}{2}x+y=\frac{-1}{3}$
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