# How do you find an equation of the line perpendicular to the graph of 15x-5y=7 that passes through the point at (0,-4)?

Mar 2, 2018

The equation of the perpendicular line is $x + 3 y = - 12$

#### Explanation:

$15 x - 5 y = 7 \mathmr{and} 5 y = 15 x - 7 \mathmr{and} y = 3 x - \frac{7}{5}$

Slope of the line $y = 3 x - \frac{7}{5} \left[y = m x + c\right]$ is ${m}_{1} = 3$

The product of slopes of the perpendicular lines is ${m}_{1} \cdot {m}_{2} = - 1$

$\therefore {m}_{2} = - \frac{1}{3} = - \frac{1}{3}$. The equation of the perpendicular line

passing through $\left(0 , - 4\right)$ having slope of ${m}_{2} = - \frac{1}{3}$ is

$y - {y}_{1} = {m}_{2} \left(x - {x}_{1}\right) \mathmr{and} y - \left(- 4\right) = - \frac{1}{3} \left(x - 0\right)$ or

$y + 4 = - \frac{1}{3} x \mathmr{and} y = - \frac{1}{3} x - 4 \mathmr{and} x + 3 y = - 12$ [Ans]