$Areaofthetriangleformedbythelines3x^2-4xy+y^2=0,2x-y=6is$

$\begin{array}{l}Givenlines3x^2-4xy+y^2=0\\ COmparingwitha{x}^2+2hxy+b{y}^2=0\\ a=3,h=-2,b=1\\ and2x-y-6=0\\ Comparingwithlx+my+n=0\\ l=2,m=-1,n=-6\end{array}$

$\begin{array}{l}Areaof{\Delta }^{le}=\frac{n^2\sqrt{h^2-ab}}{\left|a{m}^2-2hlm+b{l}^2\right|}\\ \Rightarrow \frac{36\sqrt{4-3}}{\left|3+4\left(2\right)\left(-1\right)+1\left(4\right)\right|}\\ \Rightarrow \frac{36}{\left|7-8\right|}\Rightarrow 36\end{array}$