# How do you write the equation of a line passing through (-6, 2) and perpendicular to y=-3/5x+6?

$5 x - 3 y + 36 = 0$

#### Explanation:

The slope of line: $y = - \frac{3}{5} x + 6$ is $= - \frac{3}{5}$

Now, the slope $m$ of line perpendicular to the given line: $y = - \frac{3}{5} x + 6$

$m = - \frac{1}{- \frac{3}{5}}$

$= \frac{5}{3}$

Hence the equation of line passing through the point $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(- 6 , 2\right)$ & having slope $m = \frac{5}{3}$ is given by following formula

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - 2 = \frac{5}{3} \left(x - \left(- 6\right)\right)$

$3 y - 6 = 5 x + 30$

$5 x - 3 y + 36 = 0$