Question

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

Answer 1

Let $A\left(2+\sqrt{3},5\right)$ be the vertex of the equilateral triangle ABC and x − y = 0 be the equation of BC.


Here, we have to find the equations of sides AB and AC, each of which makes an angle of ${60}^{\circ }$ with the line x − y = 0

We know the equations of two lines passing through a point $\left(x_1,y_1\right)$ and making an angle $\mathrm{\alpha }$ with the line whose slope is m.

$y-y_1=\frac{m\pm \tan \alpha }{1\mp m\tan \alpha }\left(x-x_1\right)$

Here,
$x_1=2+\sqrt{3},y_1=5,\mathrm{\alpha }={60}^{\circ },m=1$

So, the equations of the required sides are

$$\begin{gathered} y-5=\frac{1{\displaystyle +\tan {60}^{\circ }}}{{\displaystyle 1-\tan {60}^{\circ }}}\left(x-2-\sqrt{3}\right)\mathrm{and}y-5=\frac{1-{\displaystyle \tan {60}^{\circ }}}{{\displaystyle 1+\tan {60}^{\circ }}}\left(x-2-\sqrt{3}\right) \\ \Rightarrow y-5=-\left(2+\sqrt{3}\right)\left(x-2-\sqrt{3}\right)\mathrm{and}y-5=-\left(2-\sqrt{3}\right)\left(x-2-\sqrt{3}\right) \\ \Rightarrow y-5=-\left(2+\sqrt{3}\right)x+{\left(2+\sqrt{3}\right)}^2\mathrm{and}y-5=-\left(2-\sqrt{3}\right)x+\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right) \\ \Rightarrow \left(2+\sqrt{3}\right)x+y=2+4\sqrt{3}\mathrm{and}\left(2-\sqrt{3}\right)x+y-6=0 \end{gathered}$$

Hence, the second side is $y+(2-\sqrt{3})x=6$ and the equation of the third side is $\left(2+\sqrt{3}\right)x+y=12+4\sqrt{3}$