Question

A straight line passing through $P(3,1)$ meets the coordinate axes at $A$ and $B$. It is given that distance of this straight line from the origin $O$ is maximum. Area of $△OAB$ 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 origin if OP is right angled to the line drawn. Hence,
$m_{OP}=\frac{1}{3}\Rightarrow m_{AB}=-3$

Thus equation of AB is
$(y-1)=-3(x-3)$

$\Rightarrow A=(\frac{10}{3},0),B=(0,10)$

$\Rightarrow A\left(\Delta OAB\right)=\frac{1}{2}\left(OA\right)\left(OB\right)=\frac{1}{2}\times \frac{10}{3}\times 10$

$=\frac{100}{6}=\frac{50}{3}sq.units.$