Answer
See proof
Work Step by Step
As we know, two lines are the same if they have parallel vectors and a common point. Since:
\begin{align*}
L_1 : \begin{cases}
x = 3 - t \\
y = 1 + 2t\\
\end{cases} \qquad \text{has director vector} \quad \vec{v}_1 = \langle -1, 2 \rangle \\
L_2 : \begin{cases}
x = -1 + 3t \\
y = 9 - 6t
\end{cases} \qquad \text{has director vector} \quad \vec{v}_2 = \langle 3, -6 \rangle
\end{align*}
Because $\vec{v}_1 = \frac{1}{3}\vec{v}_2$, $v_1$ and $v_2$ are parallel. Furthermore, for $t = 0$ in $L_1$, we have:
\begin{align*}
x = 3, \qquad y = 1
\end{align*}
Using these values for $(x,y)$ in $L_2$:
\begin{align*}
3 &= -1 + 3t \Rightarrow t = \frac{4}{3}\\
1&= 9 - 6t \
\Rightarrow t = \frac{4}{3}
\end{align*}
We have in both equations the same value for the parameter $t$. Then $(3,1)$ is common for both lines.
Finally, $L_1$ and $L_2$ are the same line because they have parallel director vectors and a common point $(3,1)$. Therefore, we can say that the two lines coincide with each other.
