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

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$(\frac{7}{3}, \frac{- 2}{3})$

$(\frac{1}{3}, \frac{1}{3})$

$(1, 1)$

$(\frac{\sqrt{2}}{3}, \frac{\sqrt{2}}{3})$

answer is C.

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Detailed Solution

$\begin{array}{l} L e t A (2, - 1) a n d E q u a t i o n o f B C i s x + y - 2 = 0 \\ I f Δ^{l e} A B C i s E q u i l a t e r a l t h e n \\ D (h, k) = F o o t o f ⊥ r f r o m A o n B C \\ \Rightarrow \frac{h - 2}{1} = \frac{k + 1}{1} = \frac{- (2 - 1 - 2)}{1 + 1} \\ \Rightarrow \frac{h - 2}{1} = \frac{k + 1}{1} = \frac{1}{2} \\ h - 2 = \frac{1}{2}, k + 1 = \frac{1}{2} \\ h = \frac{1}{2} + 2, k = \frac{1}{2} - 1 \\ \Rightarrow \frac{5}{2}, k = - \frac{1}{2} \\ D (\frac{5}{2}, - \frac{1}{2}) \end{array}$

Orthocenter is becomes to centroide

$\begin{array}{l} ∴ G d i v i d e s A (2, - 1) a n d D (\frac{5}{2}, - \frac{1}{2}) i n t h e r a t i o 2 : 1 = m : n \\ H = G = (\frac{2 (\frac{5}{2}) + 1 (2)}{2 + 1}, \frac{2 (- \frac{1}{2}) + 1 (- 1)}{2 + 1}) \\ = (\frac{7}{3}, \frac{- 2}{3}) \end{array}$