• # question_answer Area of the triangle formed by the lines ${{y}^{2}}-9xy+18{{x}^{2}}=0$ and $y=9$ is A)            $\frac{27}{4}sq$. units   B)            $27sq.$units C)            $\frac{27}{2}sq.$ units   D)            None of these

The lines represented by ${{y}^{2}}-9xy+18{{x}^{2}}=0$ are $6x-y=0$and $3x-y=0$and third line is $y=9$.            Therefore, coordinates of vertices of triangle are given by $A(0,0)$;$B\text{ }(3,9)$ and $C\left( \frac{3}{2},9 \right)$.            Hence, area of DABC $=\frac{1}{2}\left| \begin{matrix} 0 & 0 & 1 \\ 3 & 9 & 1 \\ 3/2 & 9 & 1 \\ \end{matrix} \right|=\frac{27}{4}sq.\,units.$            Aliter: Applying the formula given in the theory part, the required area is            $\frac{{{(-9)}^{2}}\sqrt{{{(9/2)}^{2}}-18}}{18\times 1+9\times 0\times 1+1\times 0}=\frac{81}{18}\sqrt{\frac{81}{4}-18}$                                              $=\frac{81}{18}\times \frac{3}{2}=\frac{27}{4}\,sq.\ units.$