2 x^2-3 x y-2 y^2=0 represents two lines L_1 | vedclass

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(A) Given the equation $2 x^2-3 x y-2 y^2=0$,we factorize it as $(2 x+y)(x-2 y)=0$. Let $L_1: 2 x+y=0$ and $L_2: x-2 y=0$. Given the second equation $

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$\frac{3}{10}$

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(A) Given the equation $2 x^2-3 x y-2 y^2=0$,we factorize it as $(2 x+y)(x-2 y)=0$. Let $L_1: 2 x+y=0$ and $L_2: x-2 y=0$. Given the second equation $2 x^2-3 x y-2 y^2-x+7 y-3=0$,we factorize it as $(2 x+y-1)(x-2 y+3)=0$. Let $L_3: x-2 y+3=0$ and $L_4: 2 x+y-1=0$. Solving $L_1$ and $L_3$: $2 x+y=0$ and $x-2 y=-3$. Multiplying the first by $2$ gives $4 x+2 y=0$. Adding to the second gives $5 x=-3$,so $x=-\frac{3}{5}$ and $y=\frac{6}{5}$. Thus,$A = \left(-\frac{3}{5}, \frac{6}{5}ight)$. Solving $L_2$ and $L_4$: $x-2 y=0$ and $2 x+y=1$. Multiplying the second by $2$ gives $4 x+2 y=2$. Adding to the first gives $5 x=2$,so $x=\frac{2}{5}$ and $y=\frac{1}{5}$. Thus,$B = \left(\frac{2}{5}, \frac{1}{5}ight)$. Solving $L_3$ and $L_4$: $x-2 y=-3$ and $2 x+y=1$. Multiplying the second by $2$ gives $4 x+2 y=2$. Adding to the first gives $5 x=-1$,so $x=-\frac{1}{5}$ and $y=\frac{7}{5}$. Thus,$C = \left(-\frac{1}{5}, \frac{7}{5}ight)$. Now,the area of the triangle formed by $A, B, C$ is $\frac{1}{2} |x_A(y_B-y_C) + x_B(y_C-y_A) + x_C(y_A-y_B)|$. Area $= \frac{1}{2} |-\frac{3}{5}(\frac{1}{5}-\frac{7}{5}) + \frac{2}{5}(\frac{7}{5}-\frac{6}{5}) - \frac{1}{5}(\frac{6}{5}-\frac{1}{5})| = \frac{1}{2} |-\frac{3}{5}(-\frac{6}{5}) + \frac{2}{5}(\frac{1}{5}) - \frac{1}{5}(\frac{5}{5})| = \frac{1}{2} |\frac{18}{25} + \frac{2}{25} - \frac{5}{25}| = \frac{1}{2} |\frac{15}{25}| = \frac{1}{2} 	imes \frac{3}{5} = \frac{3}{10}$.

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