$O r t h o c e n t r e o f t r i a n g l e f o r m e d b y t h e l i n e s x^{2} - x y - x + y = 0 a n d 2 x - y + 4 = 0 i s$

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$(11, 4)$

$(- 11, 4)$

$(11, - 4)$

$(- 11, - 4)$

answer is B.

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Detailed Solution

$\begin{array}{l} G i v e n x^{2} - x y - x + y = 0 \\ \Rightarrow x (x - y) - 1 (x - y) = 0 \\ \Rightarrow (x - y) (x - 1) = 0 \\ \Rightarrow x - y = 0 - - (1), x - 1 = 0 - - (2) \\ a n d 2 x - y = 4 = 0 - - (3) \\ f r o m (2) x = 1, f r o m (1) x = y \Rightarrow y = 1 \\ P . I o f (1) & (2) i s (1, 1) \\ p u t x = 1 i n (3) 2 - y = 4 = 0 \Rightarrow y = 6 \\ P u t y = x i n (3) \Rightarrow 2 x - x + 4 = 0 \Rightarrow x = - 4 \end{array}$

$\begin{array}{l} ∴ P . I o f (3) & (4) i s (- 4, - 4) \\ L e t A (1, 1) B (1, 6) C (- 4, - 4) \\ s l o p e o f \bar{B C} = \frac{- 4 - 6}{- 4 - 1} = 2 \\ s l o p e o f \bar{A D} = \frac{- 1}{2} \\ E q u a t i o n o f \bar{A D} i s y - 1 = \frac{- 1}{2} (x - 1) \\ \Rightarrow 2 y - 2 = - x + 1 \\ \Rightarrow x + 2 y - 3 = 0 - - (4) \\ s l o p e o f \bar{A C} = \frac{- 4 - 1}{- 4 - 1} = 1 \\ s l o p e o f B E = - 1 \end{array}$

$\begin{array}{l} E q u a t i o n o f B E i s y - 6 = - 1 (x - 1) \\ \Rightarrow y - 6 = - x + 1 \\ \Rightarrow x + y - 7 = 0 - - (5) \\ s o l v i n g (4) & (5) \\ \begin{matrix} 2 & - 3 & 1 & 2 \\ 1 & - 7 & 1 & 1 \end{matrix} \\ \Rightarrow \frac{x}{- 14 + 3} = \frac{y}{- 3 + 7} = \frac{1}{1 - 2} \\ \Rightarrow \frac{x}{- 11} = \frac{y}{4} = \frac{1}{- 1} \\ ∴ (x, y) = (11, - 4) \end{array}$