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Cross ratio

Impo

New member
Apr 2, 2013
17
Hi,

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!

Cheers from Italy,
Impo!
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Hi,

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.
Yes, the answer $y = \frac23x$ is correct. I think that you could have simplified the calculation a bit by taking $x_1 = x_2 = x_3 = x_4 = 1$, so that you are then looking at the points where the four lines meet the line $x=1$. If you call those four points $P_1,\,P_2,\,P_3,\,P_4$, then the distances from $P_3$ to $P_1$ and $P_2$ are in the ratio $2:1$. So the distances from $P_4$ to $P_1$ and $P_2$ must be in the same ratio,namely $P_4$ must be twice as far from $P_1$ as it is from $P_2.$ Since the $y$-coordinates of $P_1$ and $P_2$ are $0$ and $1$, you see that the $y$-coordinate of $P_4$ must be $2/3$.

For a geometric construction of the harmonic conjugate of points on a line, see here.