The base $A B$ of an equilateral triangle $A B C$ with side 8 cm lies along the $y$-axis such that the mid-point of $A B$ is at the origin and $B$ lies above the origin. What is the equation of line passing through $(8,0)$ and parallel to the side $A C$ ?

Concept: In an equilateral triangle, all sides are equal and each angle $=60^{\circ}$ The slope of two parallel lines are equal The slope of a line inclined to $x$-axis at an angle $\theta$ in positive direction $=\tan \theta$ The equation of the line with slope $m$ and passing through $(a, b)$ is $y-b=m(x-a)$ - $\tan \left(180^{\circ}+\mathrm{x}\right)=\tan \mathrm{x}$ Calculation: Given: The base $A B$ of an equilateral triangle $A B C$ with side 8 cm lies along the $y$ axis such that the mid-point of $A B$ is at the origin and $B$ lies above the origin.
From figure, it can be seen that $A C$ makes an angle of $(180+30)=210^{\circ}$ $\begin{aligned} & \Rightarrow \text { slope of } A C=\tan 210^{\circ}=\tan (180+30)=\tan 30^{\circ}=\frac{1}{\sqrt{3}} \\ & \Rightarrow \text { slope of the line parallel to } A C=\frac{1}{\sqrt{3}} \end{aligned}$ $\Rightarrow$ The equation of line passing through $(8,0)$ and parallel to side $A C$ is $\begin{aligned} & (\mathrm{y}-0)=\frac{1}{\sqrt{3}}(\mathrm{x}-8) \\ & \Rightarrow x-\sqrt{3} y-8=0 \end{aligned}$