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Two fixed straight line $OX$ and $OY$ are cut by a variable line in the points $A$ and $B$ respectively and $P$ and $Q$ are the feet of the perpendiculars drawn from $A$ and $B$ upon the lines $OBY$ and $OAX$. Show that , if $AB$ passes through a fixed point, then $PQ$ will also pass through a fixed point.

Also make clear that what they mean by variable lines and fixed points/lines, means how to use them to solve this problem.

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    $\begingroup$ This sounds like a problem from an older text: $ \ OX \ $ and $ \ OY \ $ generally means the $ \ x-$ and $ \ y-$ axes, but here I think they are supposed to be non-perpendicular. $ \ A \ $ and $ \ B \ $ are any points on each of these lines, respectively, connected by a straight line. You are to construct perpendicular lines from $ \ A \ $ to the second line and from $ \ B \ $ to the first line (that made me realize $ \ OX \ $ and $ \ OY \ $ are not the coordinate axes here. (continued) $\endgroup$ Jun 8 '16 at 5:02
  • $\begingroup$ I am presuming there is no accompanying diagram? That was often the case with many old texts from a century and more ago because it cost a lot more to create diagrams for books then it did by the middle of the last century. (I am uncertain whether the "fixed points" in the last sentence mean points on the line $ \ y \ = \ x \ $ . $\endgroup$ Jun 8 '16 at 5:03
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This is NOT a solution but a drawing to clarify what the question is asking.

enter image description here

OX and OY are two oblique axes. M is the said common fixed point, through which the variable lines (L) will pass. The solid blue line is one of the family. It cuts OX and OY at A and B respectively. Through A and B, perpendiculars (in light blue) are drawn to OBY and OAX at P and Q respectively. The line through PQ is in dotted blue.

Another two sets of lines (in red and in green) are drawn according to the same rule.

Note that the dotted lines will meet at one single point N. The question is asking - “is this true?”

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