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 \[\operatorname{x}\,cos 30{}^\circ +y\,sin 30{}^\circ \] \[=\,\,p\Rightarrow \sqrt{3}x+y=2p\] This meets the coordinate axes at \[A\left( \frac{2p}{\sqrt{3}},\,\,0 \right)\] and B(0, 2p). \[\therefore \] Hence, area of \[\Delta OAB=\frac{1}{2}\left( \frac{2p}{\sqrt{3}} \right)2p\] \[=\,\,\frac{2{{p}^{2}}}{\sqrt{3}}\] \[\because \] area of triangle is \[\frac{50}{\sqrt{3}}\] \[\therefore \,\,\frac{2{{p}^{2}}}{\sqrt{3}}=\frac{50}{\sqrt{3}}\Rightarrow \,\,p=\pm 5\] Hence the lines are \[\sqrt{3}x+y\pm 10=0.\]
You need to login to perform this action.
You will be redirected in
3 sec
