The equation of the line perpendicular to the given line that contains point $ A $ is:
$ \color{blue}{ x-3y-18=0 }$ ( General form )
$ \color{blue}{ y = \frac{ 1 }{ 3 } x - 6 } ~~~$ ( Slope y-intercept form )
Step 1: The first thing that we need to do is to find the slope of a given line.
The explicit equation of a given line is: $ y = - 3 x - 6 $
So the slope is $ m = -3 $.
Step 2: The perpendicular slope ($ m_1 $) is negative reciprocal of the slope $ m $.
$$ m_1 = - \frac{1}{m} = -\frac{ 1 }{ -3 } = \frac{ 1 }{ 3 } $$So the perpendicular line will have a slope of $ m_1 = \frac{ 1 }{ 3 } $
Step 3: Now we have a point and the slope so we can use point-slope form, which is:
$$ y - y_0 = m_1 (x - x_0) $$In this example we have: $ m_1 = \frac{ 1 }{ 3 } $ , $ x_0 = 0 $ and $ y_0 = -6 $. After substitution we have:
$$ \begin{aligned} y - y_0 =& ~ m_1 (x - x_0) \\ y - \left( -6\right) =& ~ \frac{ 1 }{ 3 } ( x - 0) \\y + 6 =& ~ \frac{ 1 }{ 3 } x + 0 \\y =& ~ \frac{ 1 }{ 3 } x + 0 -6 \\y =& ~ \frac{ 1 }{ 3 } x - 6\\ \end{aligned} $$