$2 x^2-3 x y-2 y^2=0$ represents two lines $\mathrm{L}_1$ and $\mathrm{L}_2$. $2 x^2-3 x y-2 y^2-x+7 y-3=0$…
$2 x^2-3 x y-2 y^2=0$ represents two lines $\mathrm{L}_1$ and $\mathrm{L}_2$. $2 x^2-3 x y-2 y^2-x+7 y-3=0$ represents another two lines $\mathrm{L}_3$ and $\mathrm{L}_4$. Let A be the point of intersection of lines $\mathrm{L}_1$, $L_3$ and $B$ be the point of intersection of lines $L_2$ and $L_4$. The area of the triangle formed by lines $A B$ and $L_3, L_4$ is
$\frac{3}{10}$
$\frac{3}{5}$
$\frac{15}{2}$
$\frac{5}{2}$
Solution
Given the equation
$2 x^2-3 x y-2 y^2=0 \Rightarrow(2 x+y)(x-2 y)=0$
Let $\mathrm{L}_1: 2 x+y=0, \mathrm{~L}_2: x-2 y=0$
$\begin{aligned} & \text { and } 2 x^2-3 x y-2 y^2-x+7 y-3=0 \\ & \Rightarrow(2 x+y-1)(x-2 y+3)=0\end{aligned}$
Let $\mathrm{L}_3: x-2 y+3=0, \mathrm{~L}_4: 2 x+y-1=0$
After solving $L_1$ and $L_3$, we get $A\left(\frac{-3}{5}, \frac{6}{5}\right)$
After solving $\mathrm{L}_2$ and $\mathrm{L}_4$, we get $\mathrm{B}\left(\frac{2}{5}, \frac{1}{5}\right)$.
After solving $L_3$ and $L_4$, we get $C\left(\frac{-1}{5}, \frac{7}{5}\right)$
Now, required area $=\frac{1}{2}\left|\begin{array}{ccc}\frac{-3}{5} & \frac{6}{5} & 1 \\ \frac{2}{5} & \frac{1}{5} & 1 \\ \frac{-1}{5} & \frac{7}{5} & 1\end{array}\right|$
$\begin{aligned} & =\frac{1}{2}\left\{\frac{-3}{5}\left(\frac{1}{5}-\frac{7}{5}\right)-\frac{6}{5}\left(\frac{2}{5}+\frac{1}{5}\right)+1\left(\frac{14}{25}+\frac{1}{25}\right)\right\} \\ & =\frac{1}{2}\left(\frac{18}{25}-\frac{18}{25}+\frac{15}{25}\right)=\frac{3}{10}\end{aligned}$