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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!

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It's content-free. Why not roll it back or nuke it? – Rick Decker Sep 5 '13 at 2:08
up vote 1 down vote accepted

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.

Remark: I do not know what notation is used in your course, so used standard old-fashioned notation. Your version may use equivalence classes. If so, it should not be difficult to translate.

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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)? – Bill Liu 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
Thanks for spotting the typo and telling me about it. Fixed. – André Nicolas Oct 10 '14 at 20:37

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