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Two points A and B are given. Find the set of feet of the perpendiculars dropped from the point A onto all possivle straight lines passing through the point B.

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Welcome to MSE! Do you have any thoughts on the problem and can share those? Regards – Amzoti May 24 '13 at 2:42

Let $O$ be the midpoint of $AB$ and let $AO=R$ . If $l$ is any line passing through $B$ and $C$ is the leg of the perpendicular of $A$ on $l$ then the triangle $ABC$ is right at $C$ and hence $CO=AO=BO$.

Thus $C$ is on the circle with center at $O$ and radius $R$. This shows that the locus is a subset of this circle.

Moreover, if $C'$ is any point on this circle excepting $A$ and $B$, since $AB$ is a diameter then $<C'=90^0$ which shows that $C$ is a point on the locus.

If $C'=A$ then this is the case $l=AB$ and if $C'=B$ then this is the case $l \perp AB$.

This shows that the locus is the circle with diameter $AB$.

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It seems a perpendicular from $A$ to a line through $B$ can be any length between $0$ and $|\overline{AB}|$, that is, if $L$ is the set of all possible lengths we have

$$L=\{l\mid0\le l\le|\overline{AB}|\}$$

The minimum occurs when the line is through $A$, and the maximum occurs when the line is perpendicular to the segment $\overline{AB}$. All values in between occur as the line rotates about $B$.

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So, there is only one possibility? – Sucheta May 24 '13 at 2:29

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