Question

If the equal sides $AB$ and $AC$ (each equal to $a$) of a right angled isosceles $△ABC$ be produced to $P$ and $Q$ so that $BP\cdot CQ={AB}^2$, then the line $PQ$ always passes through 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 $AP=h,AQ=k$

$\frac{x}{h}+\frac{y}{k}=1⟶\left(1\right)$

Given

$BP.CQ={AB}^2$

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

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

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

$\frac{a}{h}+\frac{a}{k}=1⟶\left(2\right)$

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