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 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
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Use direction ratios with distance to get points and then apply area formula using determinant.
Updated On: Jun 4, 2025
- \(4 - 2\sqrt{3}\)
- \(8 - 4\sqrt{3}\)
- \(4 + 2\sqrt{3}\)
- \(8 + 4\sqrt{3}\)
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The Correct Option is A
Solution and Explanation
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: \[ \text{Area} = \dfrac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \Rightarrow \text{Answer: } 4 - 2\sqrt{3} \]
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: \[ \text{Area} = \dfrac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \Rightarrow \text{Answer: } 4 - 2\sqrt{3} \]
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