# Find the equation of line perpendicular to y=-5/3x-8 and passing through (-9,-3)?

Oct 12, 2016

$3 x - 5 y + 12 = 0$

#### Explanation:

The two lines are perpendicular to each other, if the product of their slopes is $- 1$. Hence if slope of one line is given as $\frac{a}{b}$, slope of line perpendicular to it is $- \frac{b}{a}$.

As equation of one line is given in slope-intercept form as $y = - \frac{5}{3} x - 8$, its slope is $- \frac{5}{3}$.

Hence slope of line perpendicular to it is $\frac{3}{5}$.

Now equation of line passing through $\left({x}_{1} , {y}_{1}\right)$ and having slope $m$ is $\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$

Hence, equation of a line passing through $\left(- 9 , - 3\right)$ and having a slope $\frac{3}{5}$ is

$\left(y - \left(- 3\right)\right) = \frac{3}{5} \left(x - \left(- 9\right)\right)$

or $5 \left(y + 3\right) = 3 \left(x + 9\right)$

or $5 y + 15 = 3 x + 27$

or $3 x - 5 y + 12 = 0$
graph{(3x-5y+12)(y+5x/3+8)((x+9)^2+(y+3)^2-0.05)=0 [-20, 20, -10, 10]}