Question

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

• A. (a,0)
• B. (0, a)
• C. (a,a)
• D. (a, 2a)

Answer 1

The correct option is 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
$\frac{x}{h}+\frac{y}{k}=1Given,BP.CQ=A{B}^2\Rightarrow (h-a)(k-a)=a^2\Rightarrow hk-ak-ah+a^2=a^2\Rightarrow ak+ha=hk\Rightarrow \frac{a}{h}+\frac{a}{k}=1$
From line (ii), of follows that line (i) it, PQ passes through the fixed point (a,a). Hence, (c) is the correct answer.