The base of an equailateral triangle is x + y - 2 = 0 and the opposite vertex is (2,-1). Find the equation of the remaining sides

Given base of an equilateral triangle is $\mathrm{x}+\mathrm{y}-2=0\dots ..\left(1\right)\mathrm{\&}$
Opposite vertex is A(2, –1)We have to find
remaining sides AB & AC passing through A
let the equation of straight line passing through
A(2, –1) & slope ‘m’ is
$\mathrm{y}+1=\mathrm{m}\left(\mathrm{x}-2\right)\dots \dots \left(2\right)$
Since angle between (1) & (2) is 60°
$\tan \mathrm{\theta }=\left|\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|\text{ where }$
$\begin{array}{l}{\mathrm{m}}_1=\text{ slope of }\left(1\right);{\mathrm{m}}_2=\text{ slope of }\left(2\right)\\ {\mathrm{m}}_1=-1,{\mathrm{m}}_2=\mathrm{m}\\ \tan {60}^{\circ }=\left|\frac{-1-\mathrm{m}}{1+(-1)\left(\mathrm{m}\right)}\right|=\sqrt{3}=\left|\frac{\mathrm{m}+1}{1-\mathrm{m}}\right|\\ \Rightarrow \sqrt{3}|1-\mathrm{m}|=|\mathrm{m}+1|\end{array}$
S.O.B.S
$\begin{array}{l}\Rightarrow 3\left(1+{\mathrm{m}}^2-2\mathrm{m}\right)={\mathrm{m}}^2+1+2\mathrm{m}\\ \Rightarrow 3+3{\mathrm{m}}^2-6\mathrm{m}-{\mathrm{m}}^2-1-2\mathrm{m}=0\\ \Rightarrow 2{\mathrm{m}}^2-8\mathrm{m}+2=0\Rightarrow {\mathrm{m}}^2-4\mathrm{m}+1=0\\ \mathrm{m}=\frac{4\pm \sqrt{16-4}}{2}=\frac{4\pm \sqrt{12}}{2}=2\pm \sqrt{3}\end{array}$
Sub in (2)
$\mathrm{y}+1=(2\pm \sqrt{3})(\mathrm{x}-2)$