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

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$\frac{50}{3}$ sq.units

$\frac{25}{3}$ sq.units

$\frac{20}{3}$ sq.units

$\frac{100}{3}$ sq.units

answer is A.

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Detailed Solution

Line AB will be the farthest from the origin if OP is right angled to the line drawn.

$OP = \sqrt{10}$

Also, $\begin{array}{l} tanθ = \frac{1}{3} \\ OA = OPsecθ \end{array}$

$= \sqrt{10} \times \frac{\sqrt{10}}{3} = \frac{10}{3}$

$OB = OP cosecθ = \sqrt{10} \times \frac{\sqrt{10}}{1} = 10$

$∴ Area of ΔOAB = \frac{1}{2} (\frac{10}{3}) (10) = \frac{50}{3}$

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