Answer
$\left\{\begin{array}{l}
x=-1+t\\
y=\frac{4}{5}t-3
\end{array}\right.\qquad 0 \leq t \leq 5$
(sample answer)
Work Step by Step
A point-slope equation of a line passing through (a,b) is
$y-b=m(x-a)$. We can define the parameter $t$ so that $t=x-a$, so a set of parametric equations for the line could be
$\left\{\begin{array}{l}
x=a+t\\
y=mt+b
\end{array}\right.$
In this case , $(a,b)=(-1,3)$ and $m=\displaystyle \frac{1-(-3)}{4-(-1)}=\frac{4}{5}$
$\left\{\begin{array}{l}
x=-1+t\\
y=\frac{4}{5}t-3
\end{array}\right.$
$(-1,-3)$ is on the line for $t=0,$
$(4,1)$ is on the line for $t=5,$
so the line segment is parametrized with
$\left\{\begin{array}{l}
x=-1+t\\
y=\frac{4}{5}t-3
\end{array}\right.\qquad 0 \leq t \leq 5$
(sample answer)