MCQ
A straight line passing through $P(3, 1)$ meet 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
  • $\frac{50}{3} sq. units$
  • B
    $\frac{25}{3} sq. units$
  • C
    $\frac{20}{3} sq. units$
  • D
    $\frac{100}{3} sq. units$

Answer

Correct option: A.
$\frac{50}{3} sq. units$
a
Line $\mathrm{AB}$ will be farthest from the origin if $\mathrm{OP}$ is right angled to loine drawn $m_{O P}=\frac{1}{3} \Rightarrow m_{A B}=-3$

Thus, the equation of $A B$ is $(y-1)=-3(x-3)$ $\Rightarrow A=\left(\frac{10}{3}, 0\right), B=(0,10)$

$\Rightarrow \Delta O A B=\frac{1}{2}(O A)(O B)=\frac{1}{2} \times \frac{10}{3} \times 10=\frac{100}{6}=\frac{50}{3}$

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