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(D) The given pair of lines is $6 x^2+13 x y+6 y^2=0$. Factoring the quadratic expression: $6 x^2+9 x y+4 x y+6 y^2=0 \implies 3 x(2 x+3 y)+2 y(2 x+3

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$\frac{45}{8}$

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(D) The given pair of lines is $6 x^2+13 x y+6 y^2=0$. Factoring the quadratic expression: $6 x^2+9 x y+4 x y+6 y^2=0 \implies 3 x(2 x+3 y)+2 y(2 x+3 y)=0 \implies (3 x+2 y)(2 x+3 y)=0$. So,the two lines are $L_1: 3 x+2 y=0$ and $L_2: 2 x+3 y=0$. The third line is $L_3: x+2 y+3=0$. To find the vertices,we solve the systems of equations: $1$) $L_1$ and $L_2$: $(0, 0)$. $2$) $L_1$ and $L_3$: $3 x+2 y=0$ and $x+2 y=-3$. Subtracting gives $2 x=3 \implies x=\frac{3}{2}$. Then $2 y=-3 x=-\frac{9}{2} \implies y=-\frac{9}{4}$. Vertex: $(\frac{3}{2}, -\frac{9}{4})$. $3$) $L_2$ and $L_3$: $2 x+3 y=0$ and $x+2 y=-3$. From $x=-3-2 y$,substitute into $2(-3-2 y)+3 y=0 \implies -6-4 y+3 y=0 \implies y=-6$. Then $x=-3-2(-6)=9$. Vertex: $(9, -6)$. Using the area formula for vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$: $	ext{Area} = \frac{1}{2} |x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$. $	ext{Area} = \frac{1}{2} |0(-\frac{9}{4}-(-6)) + \frac{3}{2}(-6-0) + 9(0-(-\frac{9}{4}))| = \frac{1}{2} |0 - 9 + \frac{81}{4}| = \frac{1}{2} |\frac{-36+81}{4}| = \frac{1}{2} 	imes \frac{45}{4} = \frac{45}{8}$ square units.

The area (in square units) of the triangle formed by the lines $6 x^2+13 x y+6 y^2=0$ and $x+2 y+3=0$ is

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