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Question: The straight line $$\frac{x}{a}+\frac{y}{b} = 1$$ cuts the coordinate axis at A and B. A line perpendicular to AB cuts the coordinate axis at P and Q. Find locus of the point of intersection of AQ and BP.

What I did: Well as the line PQ is perpendicular to AB, the formula for that line will be $$\frac{x}{b}-\frac{y}{a} = \alpha$$ Using that you can find the coordinates of P and Q in terms of $\alpha$ and then find the gradient of the line adjoining them to points A and B. The lines turn out to be perpendicular. Then assuming a point S where the lines PB and AQ meet, you can find the coordinates of that point and thus the equation. My final answer comes out to be: $$k(b-k) = h(h-a)$$

Any problem anywhere here?

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  • $\begingroup$ If you replace $(h,k)\to(x,y)$ in your final result, it's equivalent to @georg's final statement. $\endgroup$ – Semiclassical Jul 15 '14 at 17:58
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If I understood the question precisely, then

$\frac{x}{a}+\frac{y}{b} = 1, \frac{x}{b}-\frac{y}{a} = \alpha \Rightarrow\, $coordinates$\,A(a,0), Q(0,-\alpha \cdot a); B(0,b),P(\alpha \cdot b, 0)$ and equations

$AQ:\, \frac{x}{a}-\frac{y}{\alpha \cdot a}=1, BP:\, \frac{x}{\alpha \cdot b}+\frac{y}{b}=1$

$\Rightarrow $ parametric equation of locus of the point $\begin{cases} x=\frac{\alpha(\alpha\cdot a +b)}{\alpha^2+1} \\ y=\frac{\alpha(\alpha\cdot b - a)}{\alpha^2+1}\end{cases} \Rightarrow$ exclude parameter $\,\alpha$

$\Rightarrow x^2+y^2-ax-by=0$, so locus of the point is a circle.

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