Answer
$y+7=-\dfrac{5}{3}(x+4)$
Work Step by Step
Using the properties of equality, the given linear equation, $
3x-5y=6
$ is equivalent to
\begin{array}{l}
-5y=-3x+6
\\\\
y=\dfrac{-3}{-5}x+\dfrac{6}{-5}
\\\\
y=\dfrac{3}{5}x-\dfrac{6}{5}
.\end{array}
Using $y=mx+b$ or the Slope-Intercept form where $m$ is the slope, then the slope of the given line is $
\dfrac{3}{5}
$. Since perpendicular lines have negative reciprocal slopes, then the needed linear equation has slope equal to $
-\dfrac{5}{3}
$. Since it also passes through the given point $(
-4,-7
)$, then using $y-y_1=m(x-x_1)$ or the Point-Slope form where $m$ is the slope and $(x_1,y_1)$ is a point on the line, the equation of the needed line is
\begin{array}{l}
y-(-7)=-\dfrac{5}{3}(x-(-4))
\\\\
y+7=-\dfrac{5}{3}(x+4)
.\end{array}
