## Intermediate Algebra (12th Edition)

$\bf{\text{Solution Outline:}}$ To determine if the given pair of lines, $5x-y=8$ and $5y=-x+3 ,$ 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} 5x-y=8 \\\\ -y=-5x+8 \\\\ -1(-y)=-1(-5x+8) \\\\ y=5x-8 .\end{array} Hence the slope of the line defined by the first equation is $m_1=5 .$ 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} 5y=-x+3 \\\\ y=-\dfrac{1}{5}x+\dfrac{3}{5} .\end{array} Hence the slope of the line defined by second equation is $m_2=-\dfrac{1}{5} .$ Since the product of $m_1$ and $m_2$ is equal to $-1,$ then the given lines are $\text{ perpendicular }.$