Question

A line \(L\) passes through point \(P(1, 2)\) and makes an angle of \(60^\circ\) with OX in positive direction. A and B are points on line \(L\), 4 units from P. If O is origin, then area of \(\triangle OAB\) is

• A. \(4 - 2\sqrt{3}\)
• B. \(8 - 4\sqrt{3}\)
• C. \(4 + 2\sqrt{3}\)
• D. \(8 + 4\sqrt{3}\)

Hint:

Use direction ratios with distance to get points and then apply area formula using determinant.

Answer 1

The Correct Option is A

From P(1,2), direction vector along \(60^\circ\) is \((\cos 60, \sin 60) = (\frac{1}{2}, \frac{\sqrt{3}}{2})\)
So A and B lie at distance 4 → coordinates: \[ (1 \pm 2, 2 \pm 2\sqrt{3}) = (-1, 2 - 2\sqrt{3}), (3, 2 + 2\sqrt{3}) \] Now use area of triangle using determinant or area formula: \[ \text{Area} = \dfrac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \Rightarrow \text{Answer: } 4 - 2\sqrt{3} \]