## Algebra 2 Common Core

$3x+2y=-9$
Rewrite in slope-intercept form. $-2x+3y=9$ $-2x+3y+2x=2x+9$ $3y=2x+9$ $y=\frac{2}{3}x+3$ The slope of this line is $\frac{2}{3}x+3$ which means that the slope of the line perpendicular to this will be $-\frac{3}{2}$ Therefore, the tentative equation of the line is $y=-\frac{3}{2}x+b$ Substitute $x=-1$ and $y=3$ to find the value of $b$ since we know that this line passes through the point $(-1,-3)$. $-3=-\frac{3}{2} \cdot(-1) + b\\$ $-3=\frac{3}{2}+b\\$ $-3-\frac{3}{2}=b\\$ $-\frac{6}{2}-\frac{3}{2}=b\\$ $-\frac{9}{2}=b$ Thus, the equation of the line is $y= -\frac{3}{2}x - \frac{9}{2}$. Rewrite in standard form by multiplying $2$ to both sides of the equation. $2(y) = 2\left(-\frac{3}{2}x-\frac{9}{2}\right)\\ 2y=-3x-9\\ 3x + 2y = -9$