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If the equal sides $\mathrm{AB}$ and $\mathrm{AC}$ along positive $\mathrm{x}$ and $\mathrm{y}$ axes (each equal to $\mathrm{a}$ ) of a right angled isosceles triangle $\mathrm{A B C}$ be produced to $\mathrm{P}$ and $\mathrm{Q}$ respectively so that $\mathrm{BP.CQ=AB^{2} }$ , then the line $\mathrm{PQ}$ always passes through a fixed point
Option: 1
$\mathrm{(a, 0)}$

Option: 2
$\mathrm{\left ( 0,a \right )}$

Option: 3
$\mathrm{\left ( a,a \right )}$

Option: 4
none of these

Answers (1)

best_answer

We take $\mathrm{A}$ as the origin and $\mathrm{AB}$ and $\mathrm{AC}$ as $\mathrm{x-}$ axis and $\mathrm{y-}$ axis respectively.

$\mathrm{Let\: A P=h, A Q=k}$

Equation of the line PQ
$\mathrm{\frac{x}{h}+\frac{y}{k}=1\quad \cdots (i)}$

Given, $\mathrm{B P . C Q=A B^{2}}$

$\mathrm{\Rightarrow \quad(h-a)(k-a)=a^{2} \Rightarrow h k-a k-a h+a^{2}=a^{2} \quad or a k+h a=h k}$
$\mathrm{or \: \frac{a}{h}+\frac{a}{k}=1\quad \cdots(ii)}$

From (ii), it follows that line (i) i.e. PQ always passes through a fixed point $\mathrm{\left ( a,a \right )}$ .

Posted by

Gautam harsolia

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