ABC is an isoscels triangle in which A is (-1,0),ANgle A= 2pie/3,AB=AC and AB is along the x axis. If BC=4root3, then find the equation of the line BC.

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \mathrm{A}\left(-1,0\right) \\ \angle \mathrm{A}=\frac{2\mathrm{\pi }}{3} \\ \mathrm{AB}=\mathrm{AC} \\ \mathrm{BC}=4\sqrt{3} \\ \mathrm{Let} \mathrm{mid} \mathrm{point} \mathrm{of} \mathrm{BC}\hspace{0.17em}\mathrm{be} \mathrm{AD} \end{gathered}$$



$$\begin{gathered} As AB=AC, hence \\ \angle B=\angle C \\ Also \\ A+B+C=\pi \\ \Rightarrow \frac{2\pi }{3}+B+C=\pi \\ \Rightarrow B+B=\frac{\pi }{3} \\ \Rightarrow B=\frac{\pi }{6}=C \\ BC=4\sqrt{3} \\ \Rightarrow BD=2\sqrt{3} \\ cosB=\frac{BD}{AB} \\ \Rightarrow cos\frac{\pi }{6}=\frac{2\sqrt{3}}{AB} \\ \Rightarrow \frac{\sqrt{3}}{2}=\frac{2\sqrt{3}}{AB} \\ \Rightarrow AB=4 \\ As A is \left(-1,0\right) and B also on x-axis and AB=4, hence \\ Co-ordinates of B=\left(-1+4,0\right)=\left(3,0\right) \\ Also slope of line BC=\pi -\frac{\pi }{6}=\frac{5\pi }{6} \\ Hence equation of BC\hspace{0.17em}is \\ y-0=m\left(x-3\right) \\ y-0=tan\frac{5\pi }{6}\left(x-3\right) \\ \mathbf{\Rightarrow }\mathit{y}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\sqrt{\mathbf{3}}}\left(\mathbf{x}\mathbf{-}\mathbf{3}\right) \end{gathered}$$

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