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

$$\begin{gathered} \mathrm{Given} \mathrm{lines} \mathrm{x}+2\mathrm{y}=0 ....\left(1\right) \\ \begin{array}{l}4\mathrm{x}+3\mathrm{y}-5=0 ....\left(2\right)\\ 3\mathrm{x}+\mathrm{y}=0 ...\left(3\right)\end{array} \end{gathered}$$
Solving (1) & (2)

$$\begin{gathered} \mathrm{x} \mathrm{y} 1 \\ 2 0 1 2 \\ 3 -5 4 3 \end{gathered}$$
$\begin{array}{l}\frac{\mathrm{x}}{-10-0}=\frac{\mathrm{y}}{0+5}=\frac{1}{3-8}\\ \frac{\mathrm{x}}{-10}=\frac{\mathrm{y}}{5}=\frac{1}{-5}\end{array}$
$\begin{array}{l}\frac{\mathrm{x}}{-10}=\frac{1}{-5}\\ \Rightarrow \mathrm{x}=2\\ and \frac{\mathrm{y}}{5}=\frac{1}{-5}\\ \Rightarrow \mathrm{y}=-1\\ \therefore \mathrm{A}=(2,-1)\end{array}$
solving (2) & (3)

 $\begin{array}{cccc} & \mathrm{x} & \mathrm{y} 1 & \\ 3& -5& 4 & 3 \\ 1& 0& 3 & 1 \end{array}$
$\begin{array}{l}\frac{\mathrm{x}}{0+5}=\frac{\mathrm{y}}{-15+0}=\frac{1}{4-9}\\ \frac{\mathrm{x}}{5}=\frac{\mathrm{y}}{-15+0}=\frac{1}{4-9}\Rightarrow \frac{\mathrm{x}}{5}=\frac{\mathrm{y}}{-15}=\frac{1}{-5}\\ \frac{\mathrm{x}}{5}=\frac{1}{-5}\Rightarrow \mathrm{x}=-1\\ \frac{\mathrm{y}}{-15}=\frac{1}{-5}\Rightarrow \mathrm{y}=3\end{array}$
B = (–1, 3)
Solving (3) & (1)
$$\begin{gathered} \begin{array}{l} \mathrm{x}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \mathrm{y}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} 0 \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 2\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\end{array} \\ \Rightarrow \frac{\mathrm{x}}{0-0}=\frac{\mathrm{y}}{0-0}=\frac{1}{6-1} \\ \Rightarrow \frac{\mathrm{x}}{0}=\frac{\mathrm{y}}{0}=\frac{1}{5} \end{gathered}$$
$$\begin{gathered} \frac{\mathrm{x}}{0}=\frac{1}{5}\Rightarrow \mathrm{x}=0;\frac{\mathrm{y}}{0}=\frac{1}{5}\Rightarrow \mathrm{y}=0 \\ \begin{array}{l}\mathrm{C}(0,0)\\ \therefore \mathrm{A}=(2,-1),\mathrm{B}=(-1,3),\mathrm{C}=(0,0)\end{array} \end{gathered}$$
$\text{ Slope of }\overline{\mathrm{BC}}=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}=\frac{0-3}{0+1}=-3$
$$\begin{gathered} \overline{\mathrm{AD}}\perp \overline{\mathrm{BC}}\Rightarrow \text{ slope of }\overline{\mathrm{AD}}=\frac{-1}{\text{ slope of }\overline{\mathrm{BC}}} \\ =\frac{-3}{-3}=\frac{1}{3} \end{gathered}$$
$$\begin{gathered} \text{ Equation of }\overline{\mathrm{AD}};\mathrm{y}-{\mathrm{y}}_1=\mathrm{m}\left(\mathrm{x}-{\mathrm{x}}_1\right) \\ \mathrm{A}=(2,-1)\cdot \mathrm{m}=\frac{1}{3}\Rightarrow \mathrm{y}+1=\frac{1}{3}(\mathrm{x}-2) \end{gathered}$$
$$\begin{gathered} 3\mathrm{y}+3=\mathrm{x}-2\Rightarrow \mathrm{x}-3\mathrm{y}-5=0 ....\left(4\right) \\ \text{ Slope of }\overline{\mathrm{AC}}=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}=\frac{0+1}{0-2}=\frac{-1}{2} \end{gathered}$$
$\begin{array}{l}\overline{\mathrm{AC}}\perp \overline{\mathrm{BE}}\Rightarrow \text{ slope of }\overline{\mathrm{BE}}=\frac{-1}{\text{ slope of }\overline{\mathrm{AC}}}\\ \mathrm{m}=\frac{-1}{\frac{-1}{2}}=2\end{array}$
$$\begin{gathered} \text{ Equation of }\overline{\mathrm{BE}}=\mathrm{y}-{\mathrm{y}}_1\mathrm{m}\left(\mathrm{x}-{\mathrm{x}}_1\right) \\ \mathrm{B}=(-1,3)\cdot \mathrm{m}=2\Rightarrow \mathrm{y}-3=2(\mathrm{x}+1) \\ \mathrm{y}-3=2\mathrm{x}+2\Rightarrow 2\mathrm{x}-\mathrm{y}+5=0 ....\left(2\right) \end{gathered}$$
Solving (4) & (5)
$\begin{array}{llll}\mathrm{x}& \mathrm{y}& 1& \\ -3& -5& 1& -3\\ -1& 5& 2& -1\end{array}$
$$\begin{gathered} \frac{\mathrm{x}}{-15-5}=\frac{\mathrm{y}}{-10-5}=\frac{1}{-1+6} \\ \frac{\mathrm{x}}{-20}=\frac{\mathrm{y}}{-15}=\frac{1}{5} \\ \frac{\mathrm{x}}{-20}=\frac{1}{5}\Rightarrow \mathrm{x}=-4 \\ \frac{\mathrm{y}}{-15}=\frac{1}{5}\Rightarrow \mathrm{y}=-3 \end{gathered}$$
orthocenter = (–4, –3)