The equation of the line perpendicular to the given line that contains point $ A $ is:
$ \color{blue}{ 5x+9y+2=0 }$ ( General form )
$ \color{blue}{ y = - \frac{ 5 }{ 9 } x - \frac{ 2 }{ 9 } } ~~~$ ( Slope y-intercept form )
Step 1:The slope of a given line is $ m = \frac{ 9 }{ 5 } $.
Step 2: The perpendicular slope ($ m_1 $) is negative reciprocal of the slope $ m $.
$$ m_1 = - \frac{1}{m} = -\frac{ 1 }{ \frac{ 9 }{ 5 } } = -\frac{ 5 }{ 9 } $$So the perpendicular line will have a slope of $ m_1 = -\frac{ 5 }{ 9 } $
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{ 5 }{ 9 } $ , $ x_0 = 5 $ and $ y_0 = -3 $. After substitution we have:
$$ \begin{aligned} y - y_0 =& ~ m_1 (x - x_0) \\ y - \left( -3\right) =& ~ -\frac{ 5 }{ 9 } ( x - 5) \\y + 3 =& ~ -\frac{ 5 }{ 9 } x + \frac{ 25 }{ 9 } \\y =& ~ -\frac{ 5 }{ 9 } x + \frac{ 25 }{ 9 } -3 \\y =& ~ - \frac{ 5 }{ 9 } x - \frac{ 2 }{ 9 }\\ \end{aligned} $$