The orthocenter of the triangle formed by (1, -3), (6, 1), (4, -1) is

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

$(24, - 26)$

$(- 3, 2)$

$(4, \frac{8}{3})$

answer is D.

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Detailed Solution

Let A (1, -3) B (6, 1) C (4, -1)

$\begin{array}{l} S l o p e o f B C = \frac{- 1 - 1}{4 - 6} = \frac{- 2}{- 2} = 1 \\ A D ⊥ \bar{B C} a n d A (1, - 3) = (x_{1}, y_{1}) \\ \Rightarrow E q u a t i o n o f \bar{A D} i s \\ \Rightarrow y + 3 = \frac{- 1}{1} (x - 1) \\ \Rightarrow y + 3 = - x + 1 \Rightarrow x + y + 2 = 0 - - (1) \\ S l o p e o f \bar{A C} = \frac{- 1 + 3}{4 - 1} = \frac{2}{3} \\ \bar{B E} ⊥ \bar{A C} a n d B (6, 1) = (x_{1}, y_{1}) \end{array}$

$\begin{array}{l} E q u a t i o n o f \bar{B E} i s y - 1 = \frac{- 1}{(\frac{2}{3})} (x - 6) \\ \Rightarrow y - 1 = \frac{- 3}{2} (x - 6) \\ \Rightarrow 2 y - 2 = - 3 x + 18 \\ \Rightarrow 3 x + 2 y - 20 = 0 - - (2) \end{array}$

Solving (1) & (2)

$\begin{array}{l} x y 1 \\ \begin{matrix} 1 & 2 & 1 & 1 \\ 2 & - 20 & 3 & 2 \end{matrix} \\ \Rightarrow \frac{x}{- 20 - 4} = \frac{y}{6 + 20} = \frac{1}{2 - 3} \\ \Rightarrow \frac{x}{- 24} = \frac{y}{26} = \frac{1}{- 1} \\ ∴ x = 24, y = - 26 \\ ∴ o r t h o c e n t r e = (24, - 26) \end{array}$