Answer
$y-1=-\dfrac{3}{2}(x-3)$
Work Step by Step
Using the properties of equality, the given linear equation, $
2x-3y=4
$ is equivalent to
\begin{array}{l}
-3y=-2x+4
\\\\
y=\dfrac{-2}{-3}x+\dfrac{4}{-3}
\\\\
y=\dfrac{2}{3}x-\dfrac{4}{3}
.\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{2}{3}
$. Since perpendicular lines have negative reciprocal slopes, then the needed linear equation has slope equal to $
-\dfrac{3}{2}
$. Since it also passes through the given point $(
3,1
)$, 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-1=-\dfrac{3}{2}(x-3)
.\end{array}
