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

Published by Pearson

# Chapter 3 - Introduction to Graphing - 3.7 Point-Slope Form and Equations of Lines - 3.7 Exercise Set - Page 218: 49

#### Answer

$y-5=\dfrac{1}{2}(x-2)$

#### Work Step by Step

Using the properties of equality, the given linear equation, $x-2y=3$ is equivalent to \begin{array}{l} -2y=-x+3 \\\\ y=\dfrac{-1}{-2}x+\dfrac{3}{-2} \\\\ y=\dfrac{1}{2}x-\dfrac{3}{2} .\end{array} Using $y=mx+b$ or the Slope-Intercept form where $m$ is the slope, then the slope of the given line is $\dfrac{1}{2}$. Since parallel lines have the same slope, then the needed linear equation has the same slope and it passes through the given point $( 2,5 )$. 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-5=\dfrac{1}{2}(x-2) .\end{array}

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