Question

The equationof one side of an equilateral triangle is $x-y=0$ and one vertex is $(2+\sqrt{3},5)$. Prove that the second side is $y+(2-\sqrt{3})x=6$ and find the equation of the third side.

Answer 1

Let $c(2+\sqrt{3},5)$ be one vertex and $x=y$ be the opposite side of equilateral triangle ABC.

The other two sides makes an angle of ${60}^{\circ }$ with other two sides

slope of $x-y=0$ is 1

$\therefore y-5=\frac{1\pm tan{60}^{\circ }}{1\mp tan{60}^{\circ }}(x-2-\sqrt{3})$ and

$y-5=\frac{1-\sqrt{3}}{1+\sqrt{3}}(x-2-\sqrt{3})$ and

$y-5=(\sqrt{3}-2)(x-2-\sqrt{3})$

$y-5=(\sqrt{3}-2)(x-2-\sqrt{3})$ and

$y-5=(\sqrt{3}-2)(x-2-\sqrt{3})$

$y+(2+\sqrt{3})x=12+4\sqrt{3}$ and $y+(2-\sqrt{3})x=6$

Hence proved that $2nd$ side of $\Delta ABC$ is $y+(2-\sqrt{3})x=6$

and the 3rd side is

$y+(2+\sqrt{3})x=12+4\sqrt{3}$