question_answer

Equation of a straight line on which length of perpendicular from the origin is four units and the line makes an angle of \[{{120}^{o}}\]with the x?axis, is                                                       [MNR 1986]

A)            \[x\sqrt{3}+y+8=0\]             

B)            \[x\sqrt{3}-y=8\]

C)            \[x\sqrt{3}-y=8\]                     

D)            \[x-\sqrt{3}\ y+8=0\]

Correct Answer: A

Solution :

           Slope \[=-\sqrt{3}\]            \[\therefore \] Line is \[y=-\sqrt{3}x+c\]Þ \[\sqrt{3}x+y=c\]                    Now \[\frac{c}{2}=|4|\Rightarrow c=\pm \text{ }8\Rightarrow x\sqrt{3}+y=\pm \,8\].