Testwiki: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 $\alpha =0,u\ne v$ 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 ;) Template:Resolved Thanks! —Tamfang (talk) 17:03, 2 September 2016 (UTC)