Question

A straight line passing through $P(3,1)$ meets the coordinates axes at $A$ and $B$. It is given that distance of this straight line from the origin $O$ is maximum. Area of triangle $△OAB$ is equal to :

• A. $\frac{50}{3}$sq. units
• B. $\frac{25}{3}$sq. units
• C. $\frac{20}{3}$sq. units
• D. $\frac{100}{3}$sq. units

Answer 1

The correct option is A $\frac{50}{3}$sq. units

Line AB will be farthest from the origin if OP is right angled to loine drawn
$m_{OP}=\frac{1}{3}\Rightarrow m_{AB}=-3$
Thus, the equation of AB is $(y-1)=-3(x-3)$
$\Rightarrow A=(\frac{10}{3},0),B=(0,10)$
$\Rightarrow \Delta OAB=\frac{1}{2}\left(OA\right)\left(OB\right)=\frac{1}{2}\times \frac{10}{3}\times 10=\frac{100}{6}=\frac{50}{3}$