# How to get the intersection of two parallel lines on projective plane $\mathbb{P}^{2}$?

On a real projective plane $\mathbb{P}^2$, say we have two parallel lines; namely, $2x+y=0$ and $4x+2y+1=0$. What would be the equations of the projective lines, and how to find the point of their intersection?

Thanks a lot!

• It's content-free. Why not roll it back or nuke it? – Rick Decker Sep 5 '13 at 2:08

There are various approaches, but it is common to go to homogeneous coordinates. So the projective lines have homogeneous equations $2x+y=0$ and $4x+2y+z=0$. They meet where $z=0$ and $2x+y=0$, so at $(1,-2,0)$.
Because we are using homogeneous coordinates, each component of $(1,-2,0)$ can be multiplied by the same non-zero constant.
• Thanks! That's exactly the approach that I took! By the way, is it okay that I put something like $(2,-4,0)$ instead of $(1,-2,0)$? Or can I just simply put [(1,-2,0)] (representing the equivalence class)? – Zz'Rot Aug 29 '13 at 5:41
• You are welcome. Yes, you can use $(2,-4,0)$. The second paragraph of my answer said you can multiply $(1,-2,0)$ by any non-zero constant. Or you can use equivalence class language: $[(2,-4,0)]$ is exactly the same as $[(1,-2,0)]$. Since you are familiar with the language of equivalence classes, that is how you should give the answer. – André Nicolas Aug 29 '13 at 5:44
• @AndréNicolas, shouldn't be $z=0$? I guess $x=0$ is a typo. – Ram Oct 10 '14 at 19:46
The point of intersection can be read directly from the equations of the lines. Rewrite them in normal form as $(2,1)\cdot(x,y)=0$ and $(4,2)\cdot(x,y)=-1$. The two normals are non-zero scalar multiples of each other, so the lines are indeed parallel. By definition parallel lines intersect at their common point at infinity, while a line’s point at infinity and its direction vector are identical in homogeneous coordinates, so we can see immediately that the two lines intersect at $(-1,2,0)$.