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: 32



Work Step by Step

Perpendicular lines have slopes that are negative reciprocals of each other. The given line has a slope of $\frac{2}{3}$. This means that the slope of the line perpendicular to it is $-\frac{3}{2}$. Thus, the tentative equation of the line is $y=-\frac{3}{2}x+b$. To find the value of $b$, substitute the x and y-coordinates of $(-2, 6)$ into the tentative equation to have: $y=-\frac{3}{2}x+b \\6 = -\frac{3}{2} (-2)+b \\6=3+b \\6-3=b \\3=b$ Thus, the equation of the line parallel to the given line is $y=-\frac{3}{2}x+3$. Convert this equation to $ax+by=c$ form to have: $y=-\frac{3}{2}x+3 \\\frac{3}{2}x+y=3 \\2(\frac{3}{2}x+y)=3(2) \\3x+2y=6$
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