Question

The equation of the base of an equilateral triangle is x + y = 2 and its vertex is (2, −1). Find the length and equations of its sides.

Answer 1

Let A (2, −1) be the vertex of the equilateral triangle ABC and x + y = 2 be the equation of BC.

Here, we have to find the equations of the sides AB and  AC, each of which makes an angle of \[{60}^\circ\] with the line x + y = 2

The equations of two lines passing through point \[\left( x_1 , y_1 \right)\] and making an angle \[\alpha\] with the line whose slope is m is given below:

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

Here, 

\[x_1 = 2, y_1 = - 1, \alpha = {60}^\circ , m = - 1\]

So, the equations of the required sides are

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

\[ \Rightarrow y + 1 = \frac{- 1 + \sqrt{3}}{1 + \sqrt{3}}\left( x - 2 \right) \text { and } y + 1 = \frac{- 1 - \sqrt{3}}{1 - \sqrt{3}}\left( x - 2 \right)\]

\[ \Rightarrow y + 1 = \left( 2 - \sqrt{3} \right)\left( x - 2 \right) \text { and } y + 1 = \left( 2 + \sqrt{3} \right)\left( x - 2 \right)\]

Solving x + y = 2 and \[y + 1 = \left( 2 - \sqrt{3} \right)\left( x - 2 \right)\],  we get:

\[x = \frac{15 + \sqrt{3}}{6}, y = - \frac{3 + \sqrt{3}}{6}\]

\[\therefore B \equiv \left( \frac{15 + \sqrt{3}}{6}, - \frac{3 + \sqrt{3}}{6} \right) or C \equiv \left( \frac{15 - \sqrt{3}}{6}, - \left( \frac{3 - \sqrt{3}}{6} \right) \right)\]

\[\therefore\] AB = BC = AD = \[= \sqrt{\frac{2}{3}}\]

Equations of its sides are given below:

\[ \left( 2 - \sqrt{3} \right)x - y - 5 + 2\sqrt{3} = 0 , \left( 2 + \sqrt{3} \right)x - y - 5 - 2\sqrt{3} = 0\]