## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$y-1=-\dfrac{3}{2}(x-3)$
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}