The orthocenter of the $Δ^{l e} f o r m e d b y (0, 0), (8, 0), (4, 6) i s$

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(4, 3)

(3, 4)

( -3, 4)

(4, 8/3)

answer is A.

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Detailed Solution

Let O (0, 0) A ( 8, 0) B (4, 6)

$\begin{array}{l} m = S l o p e o f A B = \frac{6 - 0}{4 - 8} = \frac{6}{- 4} = \frac{- 3}{2} \\ A B ⊥ \bar{O D} a n d O (0, 0) \\ E q u a t i o n o f \bar{O D} \Rightarrow y - y_{1} = \frac{- 1}{m} (x - x_{1}) \\ \Rightarrow y - 0 = \frac{2}{3} (x - 0) \\ \Rightarrow 3 y = 2 x \Rightarrow 2 x - 3 y = 0 - - (1) \end{array}$

$\begin{array}{l} m = S l o p e o f \bar{O B} = \frac{6 - 0}{4 - 0} = \frac{6}{4} = \frac{3}{2} \\ \bar{A E} ⊥ \bar{O B} a n d A (8, 0) \\ E q u a t i o n o f \bar{A E} \Rightarrow y - y_{1} = \frac{- 1}{m} (x - x_{1}) \\ \Rightarrow y - 0 = \frac{- 2}{3} (x - 8) \\ \Rightarrow 3 y = - 2 x + 16 \\ \Rightarrow 2 x + 3 y - 16 = 0 - - (2) \end{array}$

Solving(1) & (2)

$\begin{array}{l} x y 1 \\ \begin{matrix} - 3 & 0 & 2 & - 3 \\ 3 & - 16 & 2 & 3 \end{matrix} \\ \Rightarrow \frac{x}{48 - 0} = \frac{y}{0 + 32} = \frac{1}{6 + 6} \\ \Rightarrow \frac{x}{48} = \frac{y}{32} = \frac{1}{12} \\ x = \frac{48}{12} \Rightarrow 4, y = \frac{32}{12} = \frac{8}{3} \\ ∴ o r t h o c e n t r e = (4, \frac{8}{3}) \end{array}$