The equation of the line perpendicular to the given line that contains point $ A $ is:
$ \color{blue}{ 7x-3y-45=0 }$ ( General form )
$ \color{blue}{ y = \frac{ 7 }{ 3 } x - 15 } ~~~$ ( Slope y-intercept form )
Step 1:The slope of a given line is $ m = -\frac{ 3 }{ 7 } $.
Step 2: The perpendicular slope ($ m_1 $) is negative reciprocal of the slope $ m $.
$$ m_1 = - \frac{1}{m} = -\frac{ 1 }{ -\frac{ 3 }{ 7 } } = \frac{ 7 }{ 3 } $$So the perpendicular line will have a slope of $ m_1 = \frac{ 7 }{ 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{ 7 }{ 3 } $ , $ x_0 = 6 $ and $ y_0 = -1 $. After substitution we have:
$$ \begin{aligned} y - y_0 =& ~ m_1 (x - x_0) \\ y - \left( -1\right) =& ~ \frac{ 7 }{ 3 } ( x - 6) \\y + 1 =& ~ \frac{ 7 }{ 3 } x -14 \\y =& ~ \frac{ 7 }{ 3 } x -14 -1 \\y =& ~ \frac{ 7 }{ 3 } x - 15\\ \end{aligned} $$