How do you find a standard form equation for the line perpendicular to line x-7y=9 and point(5,4)?

The standard form equation for the line perpendicular to line x-7y=9 and point (5,4) is given by
$7 x + y - 39 = 0$

Explanation:

Given line:
$x - 7 y = 9$
If $a x + b y + c = 0$
slope of the line is given by
$= - \frac{a}{b}$
Expressing the given line in the standard form
$x - 7 y - 9 = 0$
Comparing
a=1; b=-7; c=-9
Thus the slope of the line is
$= - \frac{1}{- 7} = \frac{1}{7}$
The normal has the slope which is the negative reciprocal of the slope of the given line
$= - 7$
Further, it is given that the normal passes through the point
$\left(5 , 4\right)$
Equation of the line passing through the point $\left(5 , 4\right)$ and having slope $- 7$ is given by
$\frac{y - 4}{x - 5} = - 7$
Simplifying
$y - 4 = - 7 \left(x - 5\right)$
$y - 4 = - 7 x + 35$
$y - 4 + 7 x - 35 = 0$
$y + 7 x - 39 = 0$
Rearranging in the standard form
$7 x + y - 39 = 0$