Answer
$y-5=-\dfrac{2}{7}(x+4)$
Work Step by Step
Using the properties of equality, the given linear equation, $
7x-2y=1
$ is equivalent to
\begin{array}{l}
-2y=-7x+1
\\\\
y=\dfrac{-7}{-2}x+\dfrac{1}{-2}
\\\\
y=\dfrac{7}{2}x-\dfrac{1}{2}
.\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{7}{2}
$. Since perpendicular lines have negative reciprocal slopes, then the needed linear equation has slope equal to $
-\dfrac{2}{7}
$. Since it also passes through the given point $(
-4,5
)$, 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-5=-\dfrac{2}{7}(x-(-4))
\\\\
y-5=-\dfrac{2}{7}(x+4)
.\end{array}
