## Intermediate Algebra (12th Edition)

$\bf{\text{Solution Outline:}}$ To determine if the given pair of lines, $2y=3x+12$ and $3y=2x-5 ,$ are parallel, perpendicular, or neither, find and compare the slopes of each line. If the slopes are the same, the lines are parallel. If the product of the slopes is $-1,$ the lines are perpendicular. $\bf{\text{Solution Details:}}$ In the form $y=mx+b$ or the Slope-Intercept Form (where $m$ is the slope), the first equation is equivalent to \begin{array}{l}\require{cancel} 2y=3x+12 \\\\ \dfrac{2y}{2}=\dfrac{3x}{2}+\dfrac{12}{2} \\\\ y=\dfrac{3}{2}x+6 .\end{array} Hence the slope of the line defined by the first equation is $m_1=\dfrac{3}{2} .$ In the form $y=mx+b$ or the Slope-Intercept Form (where $m$ is the slope), the second equation is equivalent to \begin{array}{l}\require{cancel} 3y=2x-5 \\\\ \dfrac{3y}{3}=\dfrac{2x}{3}-\dfrac{5}{3} \\\\ y=\dfrac{2}{3}x-\dfrac{5}{3} .\end{array} Hence the slope of the line defined by second equation is $m_2=\dfrac{2}{3} .$ Since neither the slopes are equal nor the product of $m_1$ and $m_2$ is equal to $-1,$ then the given lines are $\text{ neither parallel nor perpendicular }.$