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

Jun 26, 2018

Find the negative reciprocal of the slope and, using our point, write the equation in slope-intercept form.

#### Explanation:

The negative reciprocal of the slope of a line is the slope of the line perpendicular to it. If our original slope is $5$, then our new slope is $- \frac{1}{5}$.

Now, let's plug in our value and solve!

$2 = - \frac{1}{5} \left(3\right) + b$
$2 = - \frac{3}{5} + b$
$2 + \frac{3}{5} = b$
$\frac{13}{5} = b$

$y = - \frac{1}{5} x + \frac{13}{5}$

Jun 26, 2018

I got: $y = - \frac{1}{5} x + \frac{13}{5}$

#### Explanation:

Given our line:

$y = 5 x + 2$

the slope $m$ will be numerical coefficient of $x$ that is: $m = 5$
We know that the slope $m '$ of the perpendiclar to our line must be:

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

and the equation of the line with slope $m '$ and passing through our point of coordinates (${x}_{0} , {y}_{0}$) will be:

$y - {y}_{0} = m ' \left(x - {x}_{0}\right)$

in our case:

$y - 2 = - \frac{1}{5} \left(x - 3\right)$

$y = - \frac{1}{5} x + \frac{3}{5} + 2$

$y = - \frac{1}{5} x + \frac{13}{5}$