Find the centroid and area of the triangle formed by the lines
i) $12{\mathrm{x}}^2-20\mathrm{xy}+7{\mathrm{y}}^2=0,2\mathrm{x}-3\mathrm{y}+4=0$
ii) $2{\mathrm{y}}^2-\mathrm{xy}-6{\mathrm{x}}^2=0,\mathrm{x}+\mathrm{y}+4=0$

Given combined equation of $\overline{\mathrm{OA}}\text{ and }\overline{\mathrm{OB}}\text{ is }$
$\begin{array}{l}12{\mathrm{x}}^2-20\mathrm{xy}+7{\mathrm{y}}^2=0\\ \Rightarrow 12{\mathrm{x}}^2-6\mathrm{xy}-14\mathrm{xy}+7{\mathrm{y}}^2=0\\ \Rightarrow 6\mathrm{x}(2\mathrm{x}-\mathrm{y})-7\mathrm{y}(2\mathrm{x}-\mathrm{y})=0\\ \Rightarrow (6\mathrm{x}-7\mathrm{y})(2\mathrm{x}-\mathrm{y})=0\end{array}$
The lines represented by given  equation are
$\begin{array}{l}6\mathrm{x}-7\mathrm{y}=0 ...\left(1\right)\\ 2\mathrm{x}-\mathrm{y}=0 ...\left(2\right)\end{array}$
The third side $\overline{\mathrm{AB}}$ equation is
$2\mathrm{x}-3\mathrm{y}+4=0 ...\left(3\right)$
the point of intersection of (1) and (2) is O(0,0)
Point of intersection of the lines (1) and (3) is A
$$\begin{gathered} \begin{array}{r} \hspace{0.25em} \mathrm{x} \mathrm{y} 1 \\ \begin{array}{r}-7\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em} 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}6\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} -7\\ -3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} -3\end{array}\end{array} \end{gathered}$$
$\begin{array}{l}\Rightarrow \frac{\mathrm{x}}{-28-0}=\frac{\mathrm{y}}{0-24}=\frac{1}{-18+14}\\ \therefore \mathrm{x}=\frac{-28}{-4}=7;\mathrm{y}=\frac{-24}{-4}=6\\ \therefore \mathrm{A}=(7,6)\end{array}$
B is point of intersection of the lines (2) and (3)
$\begin{array}{ccccccc} & \mathrm{x}& & \mathrm{y}& & 1& \\ -1& & 0& & 2& & -1\\ -3& & 4& & 2& & -3\end{array}$
$\Rightarrow \frac{\mathrm{x}}{-4-0}=\frac{\mathrm{y}}{0-8}=\frac{1}{-6+2}$
$$\begin{gathered} \mathrm{x}=\frac{-4}{-4}=1;\mathrm{y}=\frac{-8}{-4}=2 \\ \therefore \mathrm{B}=(1,2) \end{gathered}$$
$\text{ Centroid of }\mathrm{△}\mathrm{OAB}=\left(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3}{3},\frac{{\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3}{3}\right)$
$$\begin{gathered} =\left(\frac{0+7+1}{3},\frac{0+6+2}{3}\right) \\ =\left(\frac{8}{3},\frac{8}{3}\right) \end{gathered}$$
$\therefore \text{ centroid of }\mathrm{△}\mathrm{OAB}=\left(\frac{8}{3},\frac{8}{3}\right)$
$$\begin{gathered} \text{ Area of }\mathrm{\Delta OAB}=\frac{1}{2}\left|{\mathrm{x}}_1{\mathrm{y}}_2-{\mathrm{x}}_2{\mathrm{y}}_1\right|=\frac{1}{2}|14-6| \\ = 4 \mathrm{sq}.\mathrm{units} \end{gathered}$$
$\text{ ii) given combined equation of }\overline{\mathrm{OA}}\text{ and }\overline{\mathrm{OB}}\text{ is }$
$\begin{array}{l}2{\mathrm{y}}^2-\mathrm{xy}-6{\mathrm{x}}^2=0\Rightarrow 2{\mathrm{y}}^2-4\mathrm{xy}+3\mathrm{xy}-6{\mathrm{x}}^2=0\\ \Rightarrow 2\mathrm{y}(\mathrm{y}-2\mathrm{x})+3\mathrm{x}(\mathrm{y}-2\mathrm{x})=0\\ \Rightarrow (\mathrm{y}-2\mathrm{x})(3\mathrm{x}+2\mathrm{y})=0\end{array}$
The lines represented by  given equation are
$\begin{array}{l}2\mathrm{x}-\mathrm{y}=0 ...\left(1\right)\\ 3\mathrm{x}+2\mathrm{y}=0 ...\left(2\right)\end{array}$
The third side $\overline{\mathrm{AB}}$ equation is
$\mathrm{x}+\mathrm{y}+4=0 ...\left(3\right)$
The point of intersection of (1) and (2) is O(0,0)
Point of intersection of the lines (1) and (3) is A
$$\begin{gathered} \mathrm{x} \mathrm{y} 1 \\ -1 0 2 -1 \\ 1 4 1 1 \end{gathered}$$
$\begin{array}{l}\Rightarrow \frac{\mathrm{x}}{-4-0}=\frac{\mathrm{y}}{0-8}=\frac{1}{2+1}\\ \therefore \mathrm{x}=-\frac{4}{3};\mathrm{y}=-\frac{8}{3};\mathrm{A}=\left(-\frac{4}{3},-\frac{8}{3}\right)\end{array}$
B is point of intersection of the lines (2) and (3)
$$\begin{gathered} \mathrm{x} \mathrm{y} 1 \\ 2 0 3 2 \\ 1 4 1 1 \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{\mathrm{x}}{8-0}=\frac{\mathrm{y}}{0-12}=\frac{1}{3-2} \\ \therefore \mathrm{B}=(8,-12) \end{gathered}$$
$\text{ Centroid of }\mathrm{△}\mathrm{OAB}=\left(\frac{0-4/3+8}{3},\frac{0-8/3-12}{3}\right)$
$=\left(\frac{-4+24}{9},\frac{-8-36}{9}\right)=\left(\frac{20}{9},\frac{-44}{9}\right)$
$\text{ Area of }\mathrm{△}\mathrm{OAB}=\frac{1}{2}\left|{\mathrm{x}}_1{\mathrm{y}}_2-{\mathrm{x}}_2{\mathrm{y}}_1\right|$
$\begin{array}{r}=\frac{1}{2}\left|-\frac{4}{3}\times -12-8\left(\frac{-8}{3}\right)\right|=\frac{1}{2}\left|16+\frac{64}{3}\right|\\ =\frac{1}{2}\left|\frac{48+64}{3}\right|=\frac{1}{2}\left(\frac{112}{3}\right)=\frac{56}{3}\text{ sq.units }\end{array}$