Given two lines, find the locus of points whose distance from the first line is two times the distance from the second line. I prefer a solution with Euclidean geometry. With analytic it's quite easy

  • $\begingroup$ Do the two lines intersect? If they are are parallel, it is pretty easy. $\endgroup$ – Thomas Andrews Aug 5 '14 at 14:14
  • $\begingroup$ It's pretty clear that if the two lines intersect at $O$ then if $X$ is in the locus, then any point on the line $OX$ is in the locus, by similarity. So it is a union of lines, probably two - one cutting the angle between the lines "each way." $\endgroup$ – Thomas Andrews Aug 5 '14 at 14:25

If your lines $r,s$ form an angle $\alpha<\pi$, consider a triangle $RPS$ in which $R\in r,\widehat{RPS}=\pi-\alpha$, $PR=2,PS=1$ and $PR\perp r$. By "sliding" $R$ over $r$ you can assume that $S\in s$ and hence $PS\perp s$. Now, by similarity, any point $Q$ on the $OP$ line, where $O=r\cap s$, has the property that $d(Q,r)=2\cdot d(Q,s)$. You only have to prove that no other points have this property, that is quite easy.

Sliding points


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