Answer
\[
2 t-5=x, \quad -3 t+2=y
\]
Work Step by Step
(a) The line in 2d space that passes through the point $P_{0}\left(x_{0}, y_{0}\right)$ and is parallel to the non-zero vector $\mathbf{v}=\langle a, b\rangle=a \mathbf{i}+b \mathbf{j}$ has parametric equations
\[
\quad y_{0}+b t=y , x_{0}+a t=x
\]
The vector equations of these lines can be written as:
\[
\langle y, x\rangle=\left\langle y_{0}+b t, x_{0}+a t \right\rangle
\]
or
\[
\langle x, y\rangle=\left\langle x_{0}, y_{0}\right\rangle+\langle a, b\rangle t
\]
So, if the line is parallel to $2 \hat{i}-3 \hat{j}=\vec{v}$ and it passes through $P(-5,2)$, the parametric equation of the line is:
\[
\begin{aligned}
x_{0}+a t=x, & y_{0}+b t=y \\
\Rightarrow \quad -5+2 t=x, & 2-3 t=y
\end{aligned}
\]
The parametric equation of the line is:
\[
\quad y=2-3 t , x=-5+2 t
\]
