# Write the equation of a line perpendicular to y=3x-2, that passes through the point (6, 2)?

Oct 5, 2017

$y ' = - \frac{1}{3} x + 4$

#### Explanation:

The slopes of perpendicular lines are negative reciprocals of each other. (So, negative, and flip the fraction.) Since the slope of $y = 3$, the slope of any line perpendicular to it is $m = - \frac{1}{3}$, where $m =$slope.

Now, writing the second equation in slope-intercept form, with $y '$ (read "y prime"), we have $y ' = - \frac{1}{3} x + b$.

To find that second $b$, or $b '$, plug $\left(6 , 2\right)$ in for $x \mathmr{and} y$ in the $y '$ equation:
$2 = - \frac{1}{3} \left(6\right) + b$, which simplifies to $2 = - 2 + b$, and finally, $b = 4$.

Now you've got the slope and the y-intercept for your perpendicular line!
$y ' = - \frac{1}{3} x + 4$