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(C) The given equation is $4x^2 + 4xy + y^2 - 6x - 3y - 4 = 0$. We can rewrite the quadratic part as $(2x + y)^2 - 3(2x + y) - 4 = 0$. Let $t = 2x + y

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$\sqrt{5}$ units

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(C) The given equation is $4x^2 + 4xy + y^2 - 6x - 3y - 4 = 0$. We can rewrite the quadratic part as $(2x + y)^2 - 3(2x + y) - 4 = 0$. Let $t = 2x + y$. Then the equation becomes $t^2 - 3t - 4 = 0$. Factoring the quadratic,we get $(t - 4)(t + 1) = 0$. This gives two lines: $2x + y - 4 = 0$ and $2x + y + 1 = 0$. These lines are parallel because their slopes are equal $(m = -2)$. The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$. Here,$A = 2, B = 1, C_1 = -4, C_2 = 1$. $d = \frac{|-4 - 1|}{\sqrt{2^2 + 1^2}} = \frac{|-5|}{\sqrt{5}} = \frac{5}{\sqrt{5}} = \sqrt{5}$ units.

The distance between the lines represented by the equation $4x^2 + 4xy + y^2 - 6x - 3y - 4 = 0$ is

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