Answer
$y=\dfrac{3}{4}x-\dfrac{19}{4}$
Work Step by Step
In the form $y=mx+b$, the given equation, $
3x-4y=12
,$ is equivalent to
\begin{align*}\require{cancel}
-4y&=-3x+12
\\\\
\dfrac{-4y}{-4}&=\dfrac{-3x+12}{-4}
\\\\
y&=\dfrac{-3}{-4}x+\dfrac{12}{-4}
\\\\
y&=\dfrac{3}{4}x-3
.\end{align*}
The slope of the equation above is given by $m$. Thus, the slope of the line with the equation above is $\dfrac{3}{4}$.
Since parallel lines have the same slope, then the required line has slope $m=\dfrac{3}{4}$. Using $
y-y_1=m(x-x_1)
$ or the Point-Slope Form of linear equations, with $(x_1,y_1)=(5,-1)$, then
\begin{align*}\require{cancel}
y-(-1)&=\dfrac{3}{4}(x-5)
\\\\
y+1&=\dfrac{3}{4}(x)+\dfrac{3}{4}(-5)
\\\\
y+1&=\dfrac{3}{4}x-\dfrac{15}{4}
\\\\
y&=\dfrac{3}{4}x-\dfrac{15}{4}-1
\\\\
y&=\dfrac{3}{4}x-\dfrac{15}{4}-\dfrac{4}{4}
\\\\
y&=\dfrac{3}{4}x-\dfrac{19}{4}
.\end{align*}
Hence, the equation of the line parallel to the given line and passing through $(5,-1)$ is $
y=\dfrac{3}{4}x-\dfrac{19}{4}
$.
