# Chapter 3 - Introduction to Graphing - 3.7 Point-Slope Form and Equations of Lines - 3.7 Exercise Set - Page 218: 60

$y+5=-2(x+2)$

#### 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 perpendicular lines have negative reciprocal slopes, then the needed linear equation has slope equal to $-2$. Since it also passes through the given point $( -2,-5 )$, 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-(-5)=-2(x-(-2)) \\\\ y+5=-2(x+2) .\end{array}

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