A right isosceles triangle $AOB$ ($O$ being the origin), is such that when $AO$ and $BO$ are extended to points $P$ and $Q$ the relation $2AP.BQ=AB^2$ holds. Prove that the line $PQ$ passes through a fixed point. I tried writing some equations of lines and using parametric equation, tried to get a relation in the distances. But, nothing worked out. How to proceed? Thanks.

  • $\begingroup$ AO and BO are extended so that A is between O and P and B between O and Q? $\endgroup$ – user261263 Aug 17 '15 at 12:56
  • $\begingroup$ It has not been stated in the question. $\endgroup$ – user167045 Aug 17 '15 at 12:59


For ease of calculations, let's us assume right angle at the vertex $O$ Let the vertices $A(a, 0)$ on the x-axis & $B(0, a)$ on the y-axis & similarly. let the points $P(h, 0)$ & $Q(0, k)$ Now, we have $$AP=h-a$$ $$BQ=k-a$$ $$AB=\sqrt{a^2+a^2}=a\sqrt 2$$

Now, applying the condition: $2(AP)(BQ)=(AB)^2$ $$2(h-a)(k-a)=(a\sqrt2 )^2$$ $$(h-a)(k-a)=a^2$$ $$a=\frac{hk}{h+k}$$

Now, the equation of the line: PQ $$y-0=\frac{k-0}{0-h}(x-h)$$ $$y=\frac{k}{h}(h-x)$$

  • $\begingroup$ Who said $A$ and $B$ lie on coordinate axis. Also, how does this show that the line passes through a fixed point. $\endgroup$ – user167045 Aug 18 '15 at 11:36

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