Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 3 - Introduction to Graphing - 3.7 Point-Slope Form and Equations of Lines - 3.7 Exercise Set - Page 218: 63



Work Step by Step

Using the properties of equality, the given linear equation, $ 3x-5y=6 $ is equivalent to \begin{array}{l} -5y=-3x+6 \\\\ y=\dfrac{-3}{-5}x+\dfrac{6}{-5} \\\\ y=\dfrac{3}{5}x-\dfrac{6}{5} .\end{array} Using $y=mx+b$ or the Slope-Intercept form where $m$ is the slope, then the slope of the given line is $ \dfrac{3}{5} $. Since perpendicular lines have negative reciprocal slopes, then the needed linear equation has slope equal to $ -\dfrac{5}{3} $. Since it also passes through the given point $( -4,-7 )$, then using $y-y_1=m(x-x_1)$ or the Point-Slope form where $m$ is the slope and $(x_1,y_1)$ is a point on the line, the equation of the needed line is \begin{array}{l} y-(-7)=-\dfrac{5}{3}(x-(-4)) \\\\ y+7=-\dfrac{5}{3}(x+4) .\end{array}
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