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(D) The given pair of lines is $12 x^2-20 x y+7 y^2=0$.Factoring the quadratic expression: $12 x^2-14 x y-6 x y+7 y^2=0$.This simplifies to $2 x(6 x-7

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$\frac{4}{39}$

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(D) The given pair of lines is $12 x^2-20 x y+7 y^2=0$.Factoring the quadratic expression: $12 x^2-14 x y-6 x y+7 y^2=0$.This simplifies to $2 x(6 x-7 y)-y(6 x-7 y)=0$,which gives $(2 x-y)(6 x-7 y)=0$.Thus,the two lines are $y=2 x$ and $y=\frac{6 x}{7}$.The third line is $x+y=1$,or $y=1-x$.To find the vertices,we solve the equations pairwise:$1$. Intersection of $y=2 x$ and $y=\frac{6 x}{7}$ is $(0,0)$.$2$. Intersection of $y=2 x$ and $x+y=1$: $x+2 x=1 \Rightarrow 3 x=1 \Rightarrow x=\frac{1}{3}, y=\frac{2}{3}$. Vertex is $(\frac{1}{3}, \frac{2}{3})$.$3$. Intersection of $y=\frac{6 x}{7}$ and $x+y=1$: $x+\frac{6 x}{7}=1 \Rightarrow \frac{13 x}{7}=1 \Rightarrow x=\frac{7}{13}, y=\frac{6}{13}$. Vertex is $(\frac{7}{13}, \frac{6}{13})$.The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$.Area $= \frac{1}{2} |0(\frac{2}{3}-\frac{6}{13}) + \frac{1}{3}(\frac{6}{13}-0) + \frac{7}{13}(0-\frac{2}{3})|$.Area $= \frac{1}{2} |\frac{6}{39} - \frac{14}{39}| = \frac{1}{2} |-\frac{8}{39}| = \frac{4}{39}$.

Area of the triangle bounded by the lines given by the equations $12 x^2-20 x y+7 y^2=0$ and $x+y-1=0$ is

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