1
$\begingroup$

This question already has an answer here:

I'm trying to solve the following problem:

Embed $\mathbb{R^2}$ in the projective plane $\mathbb{RP^2}$ by the map $(x,y)\rightarrow [1,x,y]$. Find the point of intersection in $\mathbb{RP^2}$ of the projective lines corresponding to the parallel lines $y = mx$ and $y = mx+c$ in $\mathbb{R^2}$.

So in the projective plane the two lines correspond to $[1,x,mx]$ and $[1,x,mx+c]$I get that somehow the point of intersection is the point at infinity, but that point would have coordinates $[0,1,m]$ and that doesn't fall on our line. So I'm not sure I understand how this works

$\endgroup$

marked as duplicate by rschwieb, Fabian, Claude Leibovici, user91500, Namaste May 1 '17 at 12:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Actually $[0,1,m]$ does fall on your lines! (Think about parametrising your lines as $[t,1,m]$ and $[t,1,m+t]$, where $t = 1/x$, so the point at infinity is the point $t = 0$.) $\endgroup$ – Kenny Wong Apr 27 '17 at 14:39
  • $\begingroup$ @KennyWong I thought there had to be a point in $\mathbb{R^2}$ mapping to a point for it to be on the line? $\endgroup$ – H_Hassan Apr 27 '17 at 14:42
  • $\begingroup$ No, the whole point of working in projective space is that the lines in projective space contain "extra points at infinity". $\endgroup$ – Kenny Wong Apr 27 '17 at 14:46
  • $\begingroup$ Yeah I get that, it's just not clear to me how this rigorously follows from the definitions $\endgroup$ – H_Hassan Apr 27 '17 at 14:48
  • $\begingroup$ @H_Hassan Coordinatization using homogenous coordinates lets you work with real points $(1,x,y)$ and ideal points $(0,x,y)$. You can use these coordinates in linear equations of the form $ax+by+c=0$ normally. You can also reexpress it as an orthogonality relationship between both types of coordinates and a $3$-vector. $\endgroup$ – rschwieb Apr 27 '17 at 14:51
1
$\begingroup$

The two lines correspond to the equations $c+mx-y=0$ and $0+mx-y=0$.

The method I was taught for finding the intersection of these is to take the cross product $\begin{bmatrix}c&m& -1\end{bmatrix}\times \begin{bmatrix}0& m& -1\end{bmatrix}=\begin{bmatrix}0&c& cm\end{bmatrix}=\begin{bmatrix}0&1& m\end{bmatrix}$ (if $c\neq 0$).

This last point is indeed an ideal point on both lines, which you can confirm with the two equations above (remember that the ideal points are of the form $[0,x,y]$ instead of the 'real' points $[1,x,y]$, but you use their $x,y$ values in the equation just the same.

$\endgroup$
  • $\begingroup$ So how do we know that the point is on both lines? $\endgroup$ – H_Hassan Apr 27 '17 at 14:42
  • $\begingroup$ @H_Hassan OK, take a look now. What I mean is that the coordinates of the ideal point (by which I mean $(1,m)$) satisfy both equations. If you think about it, $(0,1,m)$ will lie on any line with slope $m$ (which is the whole idea here.) $\endgroup$ – rschwieb Apr 27 '17 at 14:46

Not the answer you're looking for? Browse other questions tagged or ask your own question.