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(D) The given pair of lines is $8x^2-6xy+y^2=0$.Factoring the quadratic equation: $8x^2-4xy-2xy+y^2=0$ $\Rightarrow 4x(2x-y)-y(2x-y)=0$ $\Rightarrow (

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$28$

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(D) The given pair of lines is $8x^2-6xy+y^2=0$.Factoring the quadratic equation: $8x^2-4xy-2xy+y^2=0$ $\Rightarrow 4x(2x-y)-y(2x-y)=0$ $\Rightarrow (4x-y)(2x-y)=0$.Thus,the two lines are $y=4x$ and $y=2x$.The third line is $2x+3y=a$.The vertices of the triangle are the intersection points of these lines:$1$. Intersection of $y=4x$ and $y=2x$ is $O(0,0)$.$2$. Intersection of $y=4x$ and $2x+3y=a$: $2x+3(4x)=a$ $\Rightarrow 14x=a$ $\Rightarrow x=a/14, y=2a/7$. So,$A(a/14, 2a/7)$.$3$. Intersection of $y=2x$ and $2x+3y=a$: $2x+3(2x)=a$ $\Rightarrow 8x=a$ $\Rightarrow x=a/8, y=a/4$. So,$B(a/8, a/4)$.The area of the triangle with vertices $(0,0), (x_1, y_1), (x_2, y_2)$ is $\frac{1}{2}|x_1y_2-x_2y_1|$.Area $= \frac{1}{2} |(\frac{a}{14})(\frac{a}{4}) - (\frac{a}{8})(\frac{2a}{7})| = \frac{1}{2} |\frac{a^2}{56} - \frac{2a^2}{56}| = \frac{1}{2} |-\frac{a^2}{56}| = \frac{a^2}{112}$.Given area is $7$,so $\frac{a^2}{112} = 7$ $\Rightarrow a^2 = 784$ $\Rightarrow a = 28$ (taking positive value as per options).

If the area of the triangle formed by the pair of lines $8x^2-6xy+y^2=0$ and the line $2x+3y=a$ is $7$,then $a$ is equal to

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