# How do you write an equation in standard form given a line that passes through (-9,-3) and perpendicular to y=-5/7*x-9?

Jun 20, 2015

The line has equation $- \frac{7}{5} x + y = \frac{48}{5}$

#### Explanation:

we are looking for an equation of a line, that:

1. is perpendicular to $y = - \frac{5}{7} x - 9$
2. passes through #(-9,-3)

If a line $y = {a}_{1} x + {b}_{1}$ is perpendicular to $y = {a}_{2} x + {b}_{2}$ then ${a}_{1} \cdot {a}_{2} = - 1$ so we look for a number a for which $- \frac{5}{7} a = - 1$ which makes $a = \frac{7}{5}$

Now we have to find $b$ for which the line passes through $\left(- 9 , - 3\right)$. So we substitute $x = - 9$ abd $y = - 3$ in $y = \frac{7}{5} x + b$

$- 3 = \frac{7}{5} \cdot \left(- 9\right) + b$
$- 3 = - \frac{63}{5} + b$
$b = \frac{63}{5} - \frac{15}{5}$
$b = \frac{48}{5}$

So we get the equation $y = \frac{7}{5} x + \frac{48}{5}$.

Now to transfer it to th standard form we move terms containing $x$ and $y$ to the left and the free term to the right, sio we get:

$- \frac{7}{5} x + y = \frac{48}{5}$