Question

Show that the lines $\sqrt{3x}-y=2$, $\sqrt{3x}-y+1=0$ and $y=0$ are the side of an equilateral triangle.

Answer 1

We have,

Equation of line

$\sqrt{3}x-y=2.......\left(1\right)$

$\sqrt{3}x-y+1=0$

$\sqrt{3}x-y=-1......\left(2\right)$

And $y=0$

Put $y=0$ in equation (1) and (2) to,

$\sqrt{3}x=2$

$x=\frac{\sqrt{3}}{2}$

$\sqrt{3}x-0=-1$

$\Rightarrow x=\frac{-1}{\sqrt{3}}$

Then,

$(x_1,y_1)=(\frac{\sqrt{3}}{2},0)$

$(x_2,y_2)=(-\frac{\sqrt{3}}{2},0)$

Then, It is an equilateral triangle.