## Precalculus (6th Edition) Blitzer

$y+6=-3\left( x+2 \right)$ ; $y=-3x-12$
The equation of the line to which the line is perpendicular in the slope-intercept form will be: \begin{align} & 0=x-3y+9 \\ & -3y=x-9 \\ & y=\frac{1}{3}x+3 \end{align} The slope of this line is $\frac{1}{3}$; thus the slope of any line perpendicular to this line will be -3; using $m=-3$ and the point $\left( -2,\ -6 \right)$ we will rewrite the equation in point-slope form: \begin{align} & y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\ & y-\left( -6 \right)=-3\left( x-\left( -2 \right) \right) \\ & y+6=-3\left( x+2 \right) \end{align} Slope-intercept form: \begin{align} & y+6=-3x-6 \\ & y=-3x-12 \end{align} Thus, the equation of the line in point-slope form is $y+6=-3\left( x+2 \right)$ and the equation of the line in slope-intercept form is $y=-3x-12$, which passes through point $\left( -2,\ -6 \right)$ and is perpendicular to the line whose equation is $x-3y+9=0$.