Area of the triangle formed by the lines ${\mathrm{y}}^2-9\mathrm{xy}+18{\mathrm{x}}^2=0$ and y=6 is_______

we have ${\mathrm{y}}^2-9\mathrm{xy}-18{\mathrm{x}}^2=0$ 

or ${\mathrm{y}}^2-6\mathrm{xy}-3\mathrm{xy}+18{\mathrm{x}}^2=0$

or $\mathrm{y}(\mathrm{y}-6\mathrm{x})-3\mathrm{x}(\mathrm{y}-6\mathrm{x})=0$

or $\mathrm{y}-3\mathrm{x}=0$ and $\mathrm{y}-6\mathrm{x}=0$ The third line is y=6. Therefore, the area of the triangle formed by these lines

$=\frac{1}{2}|\begin{array}{c}0\\ 1\\ 2\end{array}\begin{array}{c}0\\ 6\\ 6\end{array}\begin{array}{c}1\\ 1\\ 1\end{array}|$

$=\frac{1}{2}|6-12|$

$=3{\mathrm{units}}^2$