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.

#### Solution

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 \[\alpha\] 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,

\[x_1 = 2 + \sqrt{3}, y_1 = 5, \alpha = {60}^\circ , m = 1\]

So, the equations of the required sides are

\[y - 5 = \frac{1 + \tan {60}^\circ}{1 - \tan {60}^\circ}\left( x - 2 - \sqrt{3} \right) \text { and }y - 5 = \frac{1 - \tan {60}^\circ}{1 + \tan {60}^\circ}\left( x - 2 - \sqrt{3} \right)\]

\[ \Rightarrow y - 5 = - \left( 2 + \sqrt{3} \right)\left( x - 2 - \sqrt{3} \right) \text { and } y - 5 = - \left( 2 - \sqrt{3} \right)\left( x - 2 - \sqrt{3} \right)\]

\[ \Rightarrow y - 5 = - \left( 2 + \sqrt{3} \right)x + \left( 2 + \sqrt{3} \right)^2 \text { and } y - 5 = - \left( 2 - \sqrt{3} \right)x + \left( 2 - \sqrt{3} \right)\left( 2 + \sqrt{3} \right)\]

\[ \Rightarrow \left( 2 + \sqrt{3} \right)x + y = 2 + 4\sqrt{3} \text { and } \left( 2 - \sqrt{3} \right)x + y - 6 = 0\]