A) \[x+\sqrt{3}y\pm 10=0\]
B) \[\sqrt{3}x+y\pm 10=0\]
C) \[x\pm \sqrt{3}y-10=0\]
D) None of these
Correct Answer: B
Solution :
Let p be the length of the perpendicular from the origin on the given line. Then its equation in normal form is \[x\cos {{30}^{o}}+y\sin {{30}^{o}}=p\]or \[\sqrt{3}x+y=2p\] This meets the coordinate axes at \[A\left( \frac{2p}{\sqrt{3}},0 \right)\] and \[B(0,\,2p)\]. \ Area of \[\Delta OAB=\frac{1}{2}\left( \frac{2p}{\sqrt{3}} \right)\text{ }2p=\frac{2{{p}^{2}}}{\sqrt{3}}\] By hypothesis \[\frac{2{{p}^{2}}}{\sqrt{3}}=\frac{50}{\sqrt{3}}\Rightarrow p=\pm 5\]. Hence the lines are \[\sqrt{3}x+y\pm 10=0\].
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