The equation of the line perpendicular to the given line that contains point $ A $ is:
$ \color{blue}{ x-4y+19=0 }$ ( General form )
$ \color{blue}{ y = \frac{ 1 }{ 4 } x + \frac{ 19 }{ 4 } } ~~~$ ( Slope y-intercept form )
Step 1:The slope of a given line is $ m = -4 $.
Step 2: The perpendicular slope ($ m_1 $) is negative reciprocal of the slope $ m $.
$$ m_1 = - \frac{1}{m} = -\frac{ 1 }{ -4 } = \frac{ 1 }{ 4 } $$So the perpendicular line will have a slope of $ m_1 = \frac{ 1 }{ 4 } $
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 }{ 4 } $ , $ x_0 = 5 $ and $ y_0 = 6 $. After substitution we have:
$$ \begin{aligned} y - y_0 =& ~ m_1 (x - x_0) \\ y - 6 =& ~ \frac{ 1 }{ 4 } ( x - 5) \\y -6 =& ~ \frac{ 1 }{ 4 } x -\frac{ 5 }{ 4 } \\y =& ~ \frac{ 1 }{ 4 } x -\frac{ 5 }{ 4 } + 6 \\y =& ~ \frac{ 1 }{ 4 } x + \frac{ 19 }{ 4 }\\ \end{aligned} $$