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(A) The given equation $x^2+4xy+ky^2-4x-10y+3=0$ represents a pair of straight lines. For this to be true,the determinant of the matrix must be zero:

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$8$

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(A) The given equation $x^2+4xy+ky^2-4x-10y+3=0$ represents a pair of straight lines. For this to be true,the determinant of the matrix must be zero: $\left|\begin{array}{ccc} 1 & 2 & -2 \ 2 & k & -5 \ -2 & -5 & 3 \end{array}ight| = 0$ Expanding the determinant: $1(3k-25) - 2(6-10) - 2(-10+2k) = 0$ $3k-25+8+20-4k = 0 \implies k = 3$. The equation becomes $x^2+4xy+3y^2-4x-10y+3=0$. Factoring the equation: $(x+y-3)(x+3y-1) = 0$. The lines are $x+y-3=0$ and $x+3y-1=0$. Solving for the intersection point: $x+y=3$ and $x+3y=1$. Subtracting gives $2y = -2 \implies y = -1$,then $x = 4$. The intersection point is $(4, -1)$. The distance $d$ from the origin $(0,0)$ to $(4,-1)$ is $d = \sqrt{4^2 + (-1)^2} = \sqrt{17}$,so $d^2 = 17$. The perpendicular distances from the origin to the lines $x+y-3=0$ and $x+3y-1=0$ are $d_1 = \frac{|-3|}{\sqrt{1^2+1^2}} = \frac{3}{\sqrt{2}}$ and $d_2 = \frac{|-1|}{\sqrt{1^2+3^2}} = \frac{1}{\sqrt{10}}$. The product $p = d_1 	imes d_2 = \frac{3}{\sqrt{2} 	imes \sqrt{10}} = \frac{3}{\sqrt{20}} = \frac{3}{2\sqrt{5}}$. Thus,$p^2 = \frac{9}{20}$. Finally,$d^2 - 20p^2 = 17 - 20 	imes \frac{9}{20} = 17 - 9 = 8$.

If $d$ is the distance between the point of intersection of the lines $x^2+4xy+ky^2-4x-10y+3=0$ and the origin,and $p$ is the product of the perpendicular distances from the origin to these lines,then $d^2-20p^2=$

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