question_answer

    ABC is an isosceles triangle in which A is\[(-1,0),\] \[\angle A=\frac{2\pi }{3},\text{ }AB=AC\]and AB is along the \[x-\] axis. If\[BC=4\sqrt{3},\]then the equation of the line BC is

A)  \[x+\sqrt{3}y=3\]           

B)  \[\sqrt{3}x+y=3\]

C)  \[x+y=\sqrt{3}\]             

D)  None of these

Correct Answer: A

Solution :

                \[(1+OB)\cos 30{}^\circ =2\sqrt{3}\] \[\Rightarrow \]\[1+OB+2\sqrt{3}\times \frac{2}{\sqrt{3}}=4\] \[\Rightarrow \] \[OB=3\] \[\Rightarrow \] \[B=(3,0)\]and \[m\]of \[BC=\tan 150{}^\circ =-\frac{1}{\sqrt{3}}\] So, the equation of\[BC\]is                 \[y-0=-\frac{1}{\sqrt{3}}(x-3)\] i.e.,        \[x+\sqrt{3}y=3\]