Suppose we have nine distinct points of $\mathbb{P}^2$ not all on the same line, and such that any straight line passing through two of these points also passes through a third. I want to show there exists $\alpha \in \mathbb{C}$ and a projective transformation $\tau$ taking these points to the following:
$$ [0,1,-1],[-1,0,1],[1,-1,0],[0,1,\alpha],[\alpha,0,1],[1,\alpha,0],[0,\alpha,1],[1,0,\alpha],[\alpha,1,0] $$ I have shown that some 4 of the points lie in general position, so i can take these to a subset of the above in general position. I don't know how to get the others to fall in line though (excuse the pun).
I'd appreciate hints rather than a full answer, if that's allowed.
(so far i have the points $[0,1,-1],[0,1,\alpha],[1,0,\alpha],[-1,0,1]$ for some $\alpha \neq 0, 1$, which i think are in general position)
I suggest you start by writing down which triples of points are collinear.
Once you do so, you will observe that if you choose $\alpha$ arbitrarily, then there are six pais of points which don't have a third point on the line connecting them. They fall into two triples in such a way that in each triple, any pair of points has no third collinear point in the complete set of all nine points:
$$[0,1,\alpha],[\alpha,0,1],[1,\alpha,0]\\ [0,\alpha,1],[1,0,\alpha],[\alpha,1,0]$$
You can make the three points of each triple become collinear if you choose $\alpha$ in a suitable way. So that takes care of choosing $\alpha$.
Next, you should try to work out that if you pick the right set of four points, all others are determined by these and the collinearities you wrote down, including those which depend on the right choice for $\alpha$. I haven't worked this one out myself yet, but since you asked for a hint, the first part might be enough of a hint to let you continue.