How would I prove that if I'm given the point on the line that is the closest to the origin compared to any other point on the line, it gives me the equation of the line? Visually this is true, but is there a way to show that this is true?
(I realised this after doing exercises on reciprocal mappings of complex numbers which lie on a circle through the origin).
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Natash1
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1You get a point and the normal vector. – quasi Jul 13 '17 at 11:53
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While it's true in 2 dimensions, it's false in 3 dimensions. – Lee Mosher Jul 13 '17 at 12:04
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Well, for the point $M$ of the line $\vec{l}$ that is closest to the axis - let us suppose that $(M,O)=r$ - we have that: $$\vec{OM}\perp\vec{l}$$ So, the line that goes through $M$ and is vertical to $\vec{OM}$ is, by definition, the tangent line of the circle $C(0,r)$ which is, again by definition, unique.
So, $M$ characterizes the line $\vec{l}$.
Vassilis Markos
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Let $L$ be a line not through the origin $O$.Let $P$ be the closest point on $L$ to $O.$ There is a unique point $Q$ on $L$ such that the segment $OQ$ is perpendicular to $L.$ So $ OQ^2+QP^2=OP^2,$ which implies $OQ<OP$ unless $Q=P.$ Since we cannot have $OQ<OP$ for any point $Q$ on $L,$ therefore $Q=P.$ So $L$ is the unique line through $P$ that is perpendicular to the segment $OP.$
DanielWainfleet
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