If a vertex of an equilateral triangle is the origin and the side opposite to it has the equation x + y = 1, then the orthocenter of the triangle is

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$(\frac{1}{2}, \frac{1}{2})$

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

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

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

answer is C.

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Detailed Solution

Let p ( h k) be the foot of perpendicular from ( 0, 0 ) to x + y – 1 = 0

$\begin{array}{l} \frac{h - 0}{1} = \frac{k - 0}{1} = \frac{- (0 + 0 - 1)}{1 + 1} \\ \frac{h}{1} = \frac{k}{1} = \frac{1}{2} \\ h = \frac{1}{2}, k = \frac{1}{2}, ∴ p (h k) = (\frac{1}{2}, \frac{1}{2}) \\ O (0, 0) p (\frac{1}{2}, \frac{1}{2}) \\ H = G d i v i d e s OP i n t h e r a t i o = 2 : 1 \\ o r t h o c e n t r e (H) = (\frac{2 (\frac{1}{2}) + 1 (0)}{2 + 1}, \frac{2 (\frac{1}{2}) + 1 (0)}{2 + 1}) \Rightarrow (\frac{1}{3}, \frac{1}{3}) \end{array}$