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 $(x_1,y_1)$ and making an angle $\alpha$ with the line whose slope is m is given below:

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

Here,

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

So, the equations of the required sides are

$y+1=\frac{-1+tan{60}^{\circ }}{1+tan{60}^{\circ }}(x-2)$ and $y+1=\frac{-1-tan{60}^{\circ }}{1-tan{60}^{\circ }}(x-2)$

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

Solving $x+y=2$ and $y+1=(2-\sqrt{3})(x-2)$, we get

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

$B\equiv (\frac{15+\sqrt{3}}{6},-\frac{3+\sqrt{3}}{6})$ or

$C\equiv (\frac{15-\sqrt{3}}{6},-\left(\frac{3-\sqrt{3}}{6}\right))$

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

Equations of its sides are given below:

$(2-\sqrt{3})x-y-5+2\sqrt{3}=0,(2+\sqrt{3})x-y-5-2\sqrt{3}=0$