# How do you write an equation of a line passing through (0, -9), perpendicular to 7x+6y=5?

Sep 5, 2017

Three steps: 1. find the gradient of the given line, 2. find the gradient of a line perpendicular to it, 3. find the equation of a line with that second gradient that passes through the given point. Solution:
$y = \frac{6}{7} x - 9$

#### Explanation:

Step 1 - find the gradient of the given line:

We need to rearrange it into point-slope form, $y = m x + c$ where $m$ is the gradient (slope).

$7 x + 6 y = 5$

Subtract $7 x$ from both sides:

$6 y = - 7 x + 5$

Divide both sides by $6$:

$y = - \frac{7}{6} x + \frac{5}{6}$

So the gradient of the given line is $- \frac{7}{6}$.

Step 2 - find the gradient of a line perpendicular to this line:

Call the new gradient ${m}_{1}$, then ${m}_{1} = - \frac{1}{m}$

That is, to find the gradient of a line perpendicular to another line, we change the sign and take the inverse.

In this case $- \frac{1}{m} = - \frac{1}{- \frac{7}{6}} = \frac{6}{7}$

The gradient of the new line we want is $\frac{6}{7}$.

Step 3 - find the equation of the new line:

We know the gradient, and we know that the line passes through the point $\left(0 , - 9\right)$. We can substitute those values into the point-slope form of a line equation:

$y = m x + c$

$- 9 = \frac{6}{7} \left(0\right) + c$

$c = - 9$

Therefore the overall equation of the line we want to find is:

$y = \frac{6}{7} x - 9$