Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

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

1 Answer 1

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 $x=0$ 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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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