Question

If the equal sides $AB$ and $AC$ $($ each equal to $a)$ of a right angled isoscele triangle $ABC$ be produced to $P$ and $Q$ so that $BP.CQ=A{B}^2,$ then the line $PQ$ always passes thought the fixed point

• A. $(a,0)$
• B. $(0,a)$
• C. $(a,a)$
• D. None of these

Answer 1

Answer: (C)

$(a,a)$

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

Let $AP=h,AQ=k.$

Equation of the line $PQ$ is

$\frac{x}{h}+\frac{y}{k}=1$ ...(1)

Given, $BP.CQ={AB}^2$

$\Rightarrow (h-a)(k-a)=a^2$

$\Rightarrow hk-ak-ah+a^2=a^2\Rightarrow ak+ha=hk$

$\Rightarrow \frac{a}{h}+\frac{a}{k}=1$ ...(2)

From (2),it follows that line (1) i.e. $PQ$ passes through the fixed point $(a,a).$