Find the orthocenter of the triangle whose sides $are x + 2 y = 0, 4 x + 3 y - 5 = 0; 3 x + y = 0$

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Detailed Solution

$Given lines x + 2 y = 0 . . . . (1) \begin{array}{l} 4 x + 3 y - 5 = 0 . . . . (2) \\ 3 x + y = 0 . . . (3) \end{array}$
Solving (1) & (2)
Question Image
$x y 1 2 0 1 2 3 - 5 4 3$
$\begin{array}{l} \frac{x}{- 10 - 0} = \frac{y}{0 + 5} = \frac{1}{3 - 8} \\ \frac{x}{- 10} = \frac{y}{5} = \frac{1}{- 5} \end{array}$
$\begin{array}{l} \frac{x}{- 10} = \frac{1}{- 5} \\ \Rightarrow x = 2 \\ a n d \frac{y}{5} = \frac{1}{- 5} \\ \Rightarrow y = - 1 \\ ∴ A = (2, - 1) \end{array}$
solving (2) & (3)

$\begin{array}{cccc} x & y 1 \\ 3 & - 5 & 4 & 3 \\ 1 & 0 & 3 & 1 \end{array}$
$\begin{array}{l} \frac{x}{0 + 5} = \frac{y}{- 15 + 0} = \frac{1}{4 - 9} \\ \frac{x}{5} = \frac{y}{- 15 + 0} = \frac{1}{4 - 9} \Rightarrow \frac{x}{5} = \frac{y}{- 15} = \frac{1}{- 5} \\ \frac{x}{5} = \frac{1}{- 5} \Rightarrow x = - 1 \\ \frac{y}{- 15} = \frac{1}{- 5} \Rightarrow y = 3 \end{array}$
B = (–1, 3)
Solving (3) & (1)
$\begin{array}{llllll} x y 1 \\ 1 0 3 1 \\ 2 0 1 2 \end{array} \Rightarrow \frac{x}{0 - 0} = \frac{y}{0 - 0} = \frac{1}{6 - 1} \Rightarrow \frac{x}{0} = \frac{y}{0} = \frac{1}{5}$
$\frac{x}{0} = \frac{1}{5} \Rightarrow x = 0; \frac{y}{0} = \frac{1}{5} \Rightarrow y = 0 \begin{array}{l} C (0, 0) \\ ∴ A = (2, - 1), B = (- 1, 3), C = (0, 0) \end{array}$
$Slope of \bar{BC} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{0 - 3}{0 + 1} = - 3$
$\bar{AD} ⊥ \bar{BC} \Rightarrow slope of \bar{AD} = \frac{- 1}{slope of \bar{BC}} = \frac{- 3}{- 3} = \frac{1}{3}$
$Equation of \bar{AD}; y - y_{1} = m (x - x_{1}) A = (2, - 1) \cdot m = \frac{1}{3} \Rightarrow y + 1 = \frac{1}{3} (x - 2)$
$3 y + 3 = x - 2 \Rightarrow x - 3 y - 5 = 0 . . . . (4) Slope of \bar{AC} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{0 + 1}{0 - 2} = \frac{- 1}{2}$
$\begin{array}{l} \bar{AC} ⊥ \bar{BE} \Rightarrow slope of \bar{BE} = \frac{- 1}{slope of \bar{AC}} \\ m = \frac{- 1}{\frac{- 1}{2}} = 2 \end{array}$
$Equation of \bar{BE} = y - y_{1} m (x - x_{1}) B = (- 1, 3) \cdot m = 2 \Rightarrow y - 3 = 2 (x + 1) y - 3 = 2 x + 2 \Rightarrow 2 x - y + 5 = 0 . . . . (2)$
Solving (4) & (5)
$\begin{array}{l} x & y & 1 \\ - 3 & - 5 & 1 & - 3 \\ - 1 & 5 & 2 & - 1 \end{array}$
$\frac{x}{- 15 - 5} = \frac{y}{- 10 - 5} = \frac{1}{- 1 + 6} \frac{x}{- 20} = \frac{y}{- 15} = \frac{1}{5} \frac{x}{- 20} = \frac{1}{5} \Rightarrow x = - 4 \frac{y}{- 15} = \frac{1}{5} \Rightarrow y = - 3$
orthocenter = (–4, –3)