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I think I've gone wrong with my reasoning somewere here but I'm not sure why.

We embed $\mathbb{R}$ into the projective plane by $(x,y)\to[1,x,y]$, and consider the projective lines corresponding to $y=mx$ and $y=mx+c, c\neq 0$ and find there intersection.

Each projective lines must be the projection of some $2$ dimensional subspace, so by letting x=0 and x=1 get the lines being the projections of

$span((1,0,0),(1,1,m))$ and $span((1,1,m+c),(1,0,c))$ or all vectors of the form

$(p,q,mq)$ and $(r,s,ms+cr)$

These subspaces intersect when $p=r, q=s$ so that $r=0$ and so the points of intersection are of the form

$(0,s,ms)$ which corresponds to the point in the projective plane $[0,1,m]$.

Is this correct? I'm worried that my answer is independent of $c$.

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Yes, you are correct!

The line defined by $y = mx$ in $\mathbb{R}^2$ embeds into the projective line defined by $y = mx$ in homogenous coordinates.

The line defined by $y = mx + c$ embeds into the projective line defined by $y = m x + c z$ in homogenous coordinates.

For $c\neq 0$, these two lines intersect when $y - mx = y-mx - cz = 0,$ that is $z = 0$ and $y = mx$, which describes the point $P = [0:1:m]$.

So, why is it independent of $c$? Well, the two lines you started with in the real spline are parallel to each other. Hence, their intersection in $\mathbb{P}^2$ will be in the line at infinity (i.e., the first coordinate of $P$ is $0$). But, the points in the line at infinity in $\mathbb{P}^2$ corresponds to different slopes of the real lines in $\mathbb{R}^2$. That is, parallel lines pass through the same point at infinity, and the constant $c$ does not affect which point that is.

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  • $\begingroup$ Thanks, that's a good explanation. $\endgroup$ – Tim Dec 13 '14 at 0:25

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