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  1. The points $P = (1: -2: 3), Q = (2: 2: -1)$ and $R = (3: 0: 2)$ lie on one Line g in $\mathbb{P}^{2}\mathbb{R}$. Choose a coordinate for g such that ${(P, Q, R)}$ has coordinates in ${(0, 1, ∞)}$.
  2. How many possibilities are there?
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  • $\begingroup$ Use LaTeX please. $\endgroup$ – Michael Rozenberg Jun 14 at 17:19
  • $\begingroup$ This question looks suspiciously like a request for us to do your homework for you. That doesn’t really meet the standard for questions on this site (see How To Ask A Good Question) and it will likely get closed or put on hold. Please update your question with your own attempts and thoughts on this problem. Where are you getting stuck? $\endgroup$ – amd Jun 14 at 17:30
  • $\begingroup$ I do not know where to start. I think i may have to use a cross ratio. But i would need a 4th point for that. And the answer should look like ${g=:((x:y:z)\in\mathbb{P}^{2}\mathbb{R}:(Ax:By:Cz=0)\cup(0:1:\infty)}$ $\endgroup$ – Semi Jun 14 at 17:36
  • $\begingroup$ Yes, that’s the right thing to be thinking about. The fourth point is an arbitrary point on the line. Use cross-ratios to parameterize the line so that $P$ is the point that corresponds to parameter value $0$, and so on. $\endgroup$ – amd Jun 14 at 17:38
  • $\begingroup$ After all, I'm already on the right path I guess. I'll try that. Thank you :) $\endgroup$ – Semi Jun 14 at 17:40
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As you suspect, cross-ratios are a good way to go here. Let $S(t)$ be some convenient parameterization of the line, $O=(0:0:1)$ a point not on the line, and $\mu$ the coordinate. You then have $${[O,S(t),Q][P,S(t),R]\over[O,S,R][P,S,Q]} = {\begin{vmatrix}\mu & 1 \\ 1 & 1\end{vmatrix} \begin{vmatrix}0&1\\1&0\end{vmatrix} \over \begin{vmatrix}\mu & 1 \\ 1 & 0\end{vmatrix} \begin{vmatrix}0&1\\1&1\end{vmatrix}}.$$ Solve for $t$ in terms of $\mu$ and substitute into $S(t)$.

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  • $\begingroup$ Alternatively, you could find a matrix $M = [aR,bP]$ such that $(M(1,1)^T)\times Q=0$, but I suspect that you’re meant to use cross-ratios. $\endgroup$ – amd Jun 15 at 0:11
  • $\begingroup$ Is there a way to calculate this without determinants? $\endgroup$ – Semi Jun 16 at 13:52
  • $\begingroup$ @Semi You could certainly use segment lengths along the given line to develop a cross-ratio along it, but can’t think of any practical way to compute $[x,\mathbf 0;\mathbf 1,\mathbf\infty]$ without determinants. $\endgroup$ – amd Jun 17 at 1:56

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