#### Answer

$\text{a) Slope-Intercept Form: }
y=-5x+19
\\\\\text{b) Stadard Form: }
5x+y=19$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To find the equation of the line with the given characeristics:
\begin{array}{l}\require{cancel}
\text{through }
(4,-1)
\\
m=-5
,\end{array}
use the Point-Slope Form of linear equations. Express the answer in Slope-Intercept Form and in the Standard Form.
$\bf{\text{Solution Details:}}$
Using $y-y_1=m(x-x_1)$ or the Point-Slope Form of linear equations, the equation of the line with the given conditions,
\begin{array}{l}\require{cancel}
y_1=-1
,\\x_1=4
,\\m=-5
,\end{array}
is
\begin{array}{l}\require{cancel}
y-y_1=m(x-x_1)
\\\\
y-(-1)=-5(x-4)
\\\\
y+1=-5(x-4)
.\end{array}
In the form $y=mx+b$ or the Slope-Intercept Form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
y+1=-5(x-4)
\\\\
y+1=-5(x)-5(-4)
\\\\
y+1=-5x+20
\\\\
y=-5x+20-1
\\\\
y=-5x+19
.\end{array}
In the form $ax+by=c$ or the Standard Form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
y=-5x+19
\\\\
5x+y=19
.\end{array}
Hence, the equation of the line is
\begin{array}{l}\require{cancel}
\text{a) Slope-Intercept Form: }
y=-5x+19
\\\\\text{b) Stadard Form: }
5x+y=19
.\end{array}