Intermediate Algebra, Chapter 3, 3.2, Section 3.2, Problem 80

State whether the lines with equation $4x - 3y = 8$ and $4y + 3x = 12$ is parallel, perpendicular, or neither.

We find the slope of each line by solving each equation for $y$

Equation 1


$
\begin{equation}
\begin{aligned}

4x - 3y =& 8
&& \text{Given equation}
\\
\\
-3y =& -4x + 8
&& \text{Subtract each side by $4x$}
\\
\\
y =& \frac{4}{3}x - \frac{8}{3}
&& \text{Divide each side by $-3$}


\end{aligned}
\end{equation}
$


Equation 2


$
\begin{equation}
\begin{aligned}

4y + 3x =& 12
&& \text{Given equation}
\\
\\
4y =& -3x + 12
&& \text{Subtract each side by $3x$}
\\
\\
y =& - \frac{3}{4}x + 3
&& \text{Divide each side by $4$}

\end{aligned}
\end{equation}
$


We know that the slope is given by the coefficient of $x$ and since the product of the slopes is $\displaystyle \frac{4}{3} \left( - \frac{3}{4} \right) = -1$ the two lines are perpendicular.