Ik had m'n vraag al op een Engelstalig forum gesteld dus hij staat hier ook in het Engels, in het Nederlands reageren mag altijd.
Let
\(l_1: y = 0, l_2: y=x, l_3: y=2x\)
. Find the equation of \(l_4\)
(through the origin) such that \(\{l_1,l_2; l_3,l_4\}=-1\)
. Also, make a geometric construction.What I did was the following:
Since
\(l_4\)
is a line through the origin it has to be of the form \(l_4: y=kx\)
where we need to find \(k \in \mathbb{R}\)
. Now, the cross ratio has only been defined as far as I know for points. Hence, let \(P_1 = (x_1,0) \in l_1, P_2 = (x_2,x_2) \in l_2, P_3 = (x_3,2x_3) \in l_3\)
and \(P_4 = (x_4,kx_4) \in l_4\)
then the cross ratio is defined as \(\{P_1,P_2;P_3,P_4\} = \frac{[P_1P_3][P_2P_4]}{[P_2P_3][P_1P_4]} \quad \mbox{whereby} \ [P_iP_j] = \left| \begin{array}{cc}x_i & x_j \\ y_i & y_j \end{array} \right|\)
If I did the right computation I find:\(\{P_1,P_2;P_3,P_4\} = \frac{2x_1x_3 x_2x_4(k-1)}{kx_2x_3x_4x_1}=-1 \Rightarrow k = \frac{2}{3}\)
Hence, \(l_4: y= \frac{2}{3}x\)
.My question: is the above correct? Secondly, what about the geometric construction? Geometrically, I know what the cross ratio mean though I have no idea what to do here. Anyone?
Thanks in advance!
