Wikipedia:Reference desk/Archives/Mathematics/2016 September 2

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September 2

an oblique kind of distance formula

Given the angle between two lines, and the signed distance of a point (in the same plane) from each of those lines, what is the point's distance² from their intersection? I've derived this (with difficulty) twice and mislaid my answer; I think it's

$u^{2}+v^{2}-{\frac {uv}{\cos \alpha }}$

but have failed to confirm it, and can't recall how I derived it ... Maybe you know a straightforward derivation? —Tamfang (talk) 09:12, 2 September 2016 (UTC)

It's easy to see this formula is incorrect, it should reduce to $u^{2}+v^{2}$ for $\alpha =\pi /2$, infinity for $\alpha =0,u\neq v$ and indeterminate for $\alpha =0,u=v$.

Wlog assume that the intersection is at the origin and that the first line is horizontal. The normalized equation for the second line is $-x\sin \alpha +y\cos \alpha =0$. Then letting the point be at $(x_{0},y_{0})$, we have $y_{0}=u$ and $-x_{0}\sin \alpha +y_{0}\cos \alpha =v$, giving $x_{0}={\frac {u\cos \alpha -v}{\sin \alpha }}$ so $x_{0}^{2}+y_{0}^{2}={\frac {u^{2}+v^{2}-2uv\cos \alpha }{\sin ^{2}\alpha }}$. -- Meni Rosenfeld (talk) 10:34, 2 September 2016 (UTC)

I agree it's easy to see that I had it wrong; now that I'm awake I hoped to get here in time to remove it ;)

Resolved

Thanks! —Tamfang (talk) 17:03, 2 September 2016 (UTC)