Question

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

Answer 1

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

We have to find the equations of sides $AB$ and $AC$, each of which makes an angle of ${60}^o$ with line $x-y=0$

We know that the equation of two lines passing through a point $(x,y)$ and making an angle $\theta$ with the line whose slope is $m$ is,

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

Here, $x_1=2+\sqrt{3},y_1=5,\theta ={60}^o$ and $m=1$ from the line $x-y=0$

So, the equation of the required sides are,

$y-5=\frac{1+\tan 60}{1-\tan 60}(x-2-\sqrt{3})$ and $(y-5)=\frac{1-\tan 60}{1+\tan 60}(x-2-\sqrt{3})$

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

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

$\Rightarrow (2+\sqrt{3}x)+y=12+4\sqrt{3}$ and $(2-\sqrt{3})x+y-6=0$

Hence, the equation of required second line and thired line are $y+(2-\sqrt{3})x=6$ and $(2+\sqrt{3})x+y=12+4\sqrt{3}$ respectively

$\Rightarrow$ Equation of thired line is $y+(2+\sqrt{3)}x=12+4\sqrt{3}$