Straight Lines and Pair of Straight Lines

To solve the problem of finding the area of the triangle formed by the lines through the given equations, we need to decode the intersections and geometry involved.The first given equation:$2 x^2-3 x y-2 y^2=0$can be factored as:$(2 x+y)(x-2 y)=0$This represents two intersecting lines $L_1: 2 x+y=0$ and $L_2: x-2 y=0$.The second equation:$2 x^2-3 x y-2 y^2-x+7 y-3=0$can be rewritten using trial and error for factorization as:$(2 x+y-3)(x-2 y+1)=0$yielding two more lines $L_3: x-2 y+1=0$ and $L_4: 2 x+y-3=0$.The line pairs are rewritten as:$L_1: 2 x+y=0$ (parallel to $L_4: 2 x+y-3=0$ )$L_2: x-2 y=0$ (parallel to $L_3: x-2 y+1=0$ )Find intersection $A$ between lines $L_1$ and $L_3$, and $B$ between lines $L_2$ and $L_4$.Intersection calculation:For point $A$, solve:$\;2 x+y=0$$\;x-2 y+1=0$The solution is:$A= \left(-\;\frac{1}{5}, \;\frac{2}{5}\right)$For point $B$, solve:$\;x-2 y=0$$\;2 x+y-3=0$The solution is:$B= \left(\;\frac{6}{5}, \;\frac{3}{5}\right)$Area calculation:Using the coordinates $A= \left(-\;\frac{1}{5}, \;\frac{2}{5}\right) , B= \left(\;\frac{6}{5}, \;\frac{3}{5}\right)$, and the intersection point $P(1,1)$ of lines $L_3$ and $L_4$ :The area of triangle $\Delta A B P$ is computed using the determinant method:Area $=\;\frac{1}{2} \begin{vmatrix} -\;\frac{1}{5} & \;\frac{2}{5} & 1 \\ \;\frac{6}{5} & \;\frac{3}{5} & 1 \\ 1 & 1 & 1 \end{vmatrix}$Performing column operations:Subtract the first column from the second and third:$C_2 \rightarrow C_2-C_1, \;\; C_3 \rightarrow C_3-C_1$ Thus, we have:$\;\frac{1}{2} \begin{vmatrix} -\;\frac{1}{5} & \;\frac{3}{5} & \;\frac{6}{5} \\ \;\frac{6}{5} & -\;\frac{3}{5} & -\;\frac{1}{5} \\ 1 & 0 & 0 \end{vmatrix}$Then, compute:$=\;\frac{1}{2} \times \;\frac{15}{25} = \;\frac{1}{2} \times \;\frac{3}{5} = \;\frac{3}{10}$Thus, the area of the triangle is $\;\frac{3}{10}$.