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The line $\mathrm{x}+\mathrm{y}=\mathrm{p}$ meets the axis of $\mathrm{x} \, \&\, \mathrm{y}$ at A & B respectively. A triangle $\mathrm{APQ}$ is inscribed in the triangle $\mathrm{OAB}, \mathrm{O}$ being the origin, with right angle at $\mathrm{Q}$ . $\mathrm{P}$ and $\mathrm{Q}$ lie respectively on $\mathrm{OB}$ and $\mathrm{AB}$ .If the area of the triangle $\mathrm{A P Q}$ is $\mathrm{3 / 8^{\text {th }}}$ of the area of the triangle $\mathrm{O A B}$ , then $\frac{A Q}{B Q}$ is equal to :
Option: 1
$2$

Option: 2
$2 / 3$

Option: 3
$1 / 3$

Option: 4
$3$

Answers (1)

best_answer

$\frac{\triangle \mathrm{AQP}}{\Delta \mathrm{AOB}}=\frac{3}{8} \text { or } \frac{\frac{\mathrm{p}^2 \lambda}{(\lambda+1)^2}}{\frac{1}{2} \mathrm{p}^2}=\frac{3}{8}$

$\Rightarrow \quad \lambda=3, \frac{1}{3}$
$\frac{\mathrm{AQ}}{\mathrm{BQ}}=3 \text { or } \frac{1}{3}$

$\frac{1}{3}$ is rejected because this gives negative coordinater of P and it is gives that P lies on OB.

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Rakesh

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