geometry - Show that a complex number is on a line.

Let $v$ and $w$ be two distinct complex numbers such that $v + t (w-v)$ is a line $l$, where $t \in \mathbb{R}$. Prove that:

If $\frac{z - w}{z-v}$ is a real number, for instance $t$, then $z$ is on the line $l$.

I have already tried using the hint as follows:

\begin{equation}
\frac{z - w}{z-v}=t
\end{equation}
\begin{equation}
z - w=t(z-v)
\end{equation}
\begin{equation}
z = \frac{-v t +w}{1-t}

\end{equation}

This doesn't really get me anywhere, at least I think it doesn't because I don't recognise the form of my desired line in here. I also tried using the following algorithm:
\begin{equation}
\frac{z - w}{z-v}(w-v)=t(w-v)
\end{equation}
\begin{equation}
\frac{z - w}{z-v}(w-v)+v=v+ t(w-v)
\end{equation}
Which can be rewritten as:

\begin{equation}
\frac{zw-w^2 +vw-v^2}{z-v}=v+ t(w-v)
\end{equation}

I would have hoped it to simplify to $z$. Do you people have any pointers or tips?

Answer

Your relation
$$
z = \frac{-v t +w}{1-t}
$$

can be written as
$$
z=v+\tau(w-v),
$$
where $\displaystyle\tau={1\over1-t}={z-v\over w-v}$.