Answer
$\left\{\begin{array}{l}
x=2+t\\
y=\dfrac{4}{3}t+3
\end{array}\right.\qquad t \leq 0$
Work Step by Step
A point-slope equation of a line passing through (a,b) is
$y-b=m(x-a)$.
In this problem, $(a,b)=(2,3)$ and $m=\displaystyle \frac{-1-3}{-1-2}=\frac{4}{3}$
We can define the parameter $t$ so that $t=x-a$,
so a set of parametric equations is
$\left\{\begin{array}{l}
x=a+t\\
y=mt+b
\end{array}\right.$
so this line can be parametrized with
$\left\{\begin{array}{l}
x=2+t\\
y=\frac{4}{3}t+3
\end{array}\right.$
We don't want the whole line, just the ray.
$(2,3)$ is on the line for $t=0,$
$(-1,-1)$ is on the line for $t=-3,$ (negative t)
So the ray can be parametrized with:
$\left\{\begin{array}{l}
x=2+t\\
y=\frac{4}{3}t+3
\end{array}\right.\qquad t \leq 0$
