Question

ABC is an isosceles triangle in which A is $(-1,0),$ $\angle A=\frac{2\pi }{3},AB=AC$ 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

Answer 1

Answer: (A)

$(1+OB)\cos 30{}^{\circ }=2\sqrt{3}$ $\Rightarrow$ $1+OB+2\sqrt{3}\times \frac{2}{\sqrt{3}}=4$
$\Rightarrow$ $OB=3$ $\Rightarrow$ $B=(3,0)$ 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., $x+\sqrt{3}y=3$