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: 53

Answer

$y+3=-\dfrac{2}{3}(x+2)$

Work Step by Step

Using the properties of equality, the given linear equation, $ 2x+3y=-7 $ is equivalent to \begin{array}{l} 3y=-2x-7 \\\\ y=-\dfrac{2}{3}x-\dfrac{7}{3} .\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{2}{3} $. Since parallel lines have the same slope, then the needed linear equation has the same slope and it passes through the given point $( -2,-3 )$. 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, then the equation of the needed line is \begin{array}{l} y-(-3)=-\dfrac{2}{3}(x-(-2)) \\\\ y+3=-\dfrac{2}{3}(x+2) .\end{array}
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