The equation of the base of an equilateral triangle is x + y = 2 and one vertex is ( 2 , - 1 ) , then the…

Let, $∆ABC,$ be the triangle, with side $BC$ as $x+y=2,$ and vertex $A$ as $\left(2, -1\right).$

The length of perpendicular from the point $\left(x_1, y_1\right)$ to the line $ax+by+c=0$ is $\frac{\left|a{x}_1+b{y}_1+c\right|}{\sqrt{a^2+b^2}}.$

Hence, the length of altitude $AM=\left|\frac{2-1-2}{\sqrt{1^2+1^2}}\right|$

$\Rightarrow AM=\frac{1}{\sqrt{2}}.$

We know that, in an equilateral triangle, altitude, median, angle bisector all are same.

Hence, $AM$ is also the median, thus $MC=\frac{1}{2}BC=\frac{1}{2}AC.$

Now, in $∆AMC,$ by Pythagoras theorem $A{C}^2= A{M}^2+M{C}^2$

$\Rightarrow A{C}^2=A{M}^2+\frac{1}{4}A{C}^2$

$\Rightarrow \frac{3}{4}A{C}^2={\left(\frac{1}{\sqrt{2}}\right)}^2$

$\Rightarrow \frac{3}{4}A{C}^2=\frac{1}{2}$

$\Rightarrow AC=\sqrt{\frac{2}{3}}.$

Thus, the side of the equilateral triangle is $\sqrt{\frac{2}{3}}.$