Question

One of the vertex of an equilateral triangles $(2,3)$ and the equation of its opposite side is $x+y=2$. Find the equation of the remaining two sides.

Answer 1

Let, In $△ABC$, coordinate of $A$ are $(2,3)$ and equation of $BC$ is $x+y=2$.

Equation of line passes through point $A(2,3)$ and making angle of

${60}^o$ with $x+y=2$ is .

$y-k=\pm \frac{m-\tan \alpha }{1+m\tan \alpha }(x-h)$

Here $k=3,m=-1,\alpha ={60}^o,h=2$

$y-3=\pm \left(\frac{-1-\tan {60}^o}{1+(-1)\tan {60}^o}\right)(x-2)$

$\Rightarrow y-3=\pm \left(\frac{-1-\sqrt{3}}{1-\sqrt{3}}\right(x-2\left)\right)$

Taking $+$ sign

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

$\Rightarrow (y-3)(1-\sqrt{3})+x+\sqrt{3}x-2-2\sqrt{3}=0$

$\Rightarrow y-3-\sqrt{3}y+3\sqrt{3}+x+\sqrt{3}x-2-2\sqrt{3}=0$

$\Rightarrow (1-\sqrt{3})y+(1+\sqrt{3})x-5+\sqrt{3}=0$

$\Rightarrow (1+\sqrt{3})x+(1-\sqrt{3})y=(5-\sqrt{3})$

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

$\Rightarrow (y-3)(1-\sqrt{3})-(x-2)(1+\sqrt{3})=0$

$\Rightarrow y-3-\sqrt{3}y+3\sqrt{3}-x+2-\sqrt{3}x+2\sqrt{3}=0$

$\Rightarrow (1-\sqrt{3})y-(1+\sqrt{3})x-1+5\sqrt{3}=0$

$\Rightarrow -(1+\sqrt{3})x+(1-\sqrt{3})y-(1-5\sqrt{3})=0$

$\Rightarrow (1+\sqrt{3})x-(1-\sqrt{3})y=-(1-5\sqrt{3})$

$\Rightarrow (1+\sqrt{3})x-(1-\sqrt{3})y=5\sqrt{3}-1$