# How do you write an equation in standard form for a line passing through (3, -5) and is perpendicular to the line y = 3x - 6?

$y - {y}_{0} = m \left(x - {x}_{0}\right)$
Where (${x}_{0} , {y}_{0}$) are the coordinates of your point ($3 , - 5$) and $m$ is the slope.
Now, for your line to be perpendicular to the given one it needs to have a slope: $m = - \frac{1}{m} _ 1$ where ${m}_{1}$ is the slope of your given line (i.e., ${m}_{1} = 3$, the coefficient of $x$ in $y = 3 x - 6$), and $m = - \frac{1}{3}$.
$y - \left(- 5\right) = - \frac{1}{3} \left(x - 3\right)$