# Cross-ratio projective geometry

I have 4 points $P_0=[1:2], P_1=[3:4], P_2=[5:6], P_3=[7,8]$ in $\mathbb KP^1$ and would like to evaluate the cross-ratio.

It is given by the following:

$\pi:\mathbb KP^1\rightarrow G$ is the unique projective map with $\pi([1:0])=P_0, \pi([0:1])=P_1, \pi([1:1])=P_2$ and $\pi([x_0:x_1])=P_3$

The cross-ratio is euqal to $\frac{x_1}{x_0}$

My problem is, how to evaluate $x_0$ and $x_1$, may you could help me with that.

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$$P_2\,,\,P_3\in l: [1:2]+t[2:2]=P_0+t(P_1-P_0)$$
as a projective map, $\,\pi\,$ must map collinear points to collinear points...can you take it from here?
How do you check that $P_0, P_1,P_2, P_3$ are collinear points? Does that mean that $x_0=2, x_1=2$ ? –  Voyage Jan 13 '13 at 14:20
It you see the definion of $\,l\,$, you can see that $\,P_2=[1:2]+2[2:2]\;,\;\;P_3=[1:2]+3[2:2]\Longrightarrow P_1,P_2,P_3\in l\,$ and thus they're collinear. –  DonAntonio Jan 13 '13 at 22:34
Thanks, so therefore $x_0=2, x_1=2$, correct? –  Voyage Jan 13 '13 at 23:28