The vertex of an equilateral triangle is at (2, -1) and the side opposite to it has the equation x + y = 2 then orthocentre of the triangle is

$\begin{array}{l}LetA\left(2,-1\right)andEquationofBCisx+y-2=0\\ If{\Delta }^{le}ABCisEquilateralthen\\ D\left(h,k\right)=Footof\perp rfromA\text{\hspace{0.17em}}onBC\\ \Rightarrow \frac{h-2}{1}=\frac{k+1}{1}=\frac{-\left(2-1-2\right)}{1+1}\\ \Rightarrow \frac{h-2}{1}=\frac{k+1}{1}=\frac{1}{2}\\ h-2=\frac{1}{2},k+1=\frac{1}{2}\\ h=\frac{1}{2}+2,k=\frac{1}{2}-1\\ \Rightarrow \frac{5}{2},k=-\frac{1}{2}\\ D\left(\frac{5}{2},-\frac{1}{2}\right)\end{array}$

Orthocenter is becomes to centroide

$\begin{array}{l}\therefore GdividesA\left(2,-1\right)andD\left(\frac{5}{2},-\frac{1}{2}\right)intheratio2:1=m:n\\ H=G=\left(\frac{2\left(\frac{5}{2}\right)+1\left(2\right)}{2+1},\frac{2\left(-\frac{1}{2}\right)+1\left(-1\right)}{2+1}\right)\\ =\left(\frac{7}{3},\frac{-2}{3}\right)\end{array}$