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(A) The given equation is $x^2 + 2\sqrt{3}xy + 3y^2 - 3x - 3\sqrt{3}y - 4 = 0$.Comparing this with the general form $ax^2 + 2hxy + by^2 + 2gx + 2fy +

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$2.5$

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(A) The given equation is $x^2 + 2\sqrt{3}xy + 3y^2 - 3x - 3\sqrt{3}y - 4 = 0$.Comparing this with the general form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we have $a=1, h=\sqrt{3}, b=3, g=-3/2, f=-3\sqrt{3}/2, c=-4$.First,check if the lines are parallel by verifying $h^2 - ab = (\sqrt{3})^2 - (1)(3) = 3 - 3 = 0$. Since $h^2 - ab = 0$,the lines are parallel.The formula for the distance between two parallel lines $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is $d = 2\sqrt{\frac{g^2 - ac}{a(a + b)}}$.Substituting the values: $d = 2\sqrt{\frac{(-3/2)^2 - (1)(-4)}{1(1 + 3)}} = 2\sqrt{\frac{9/4 + 4}{4}} = 2\sqrt{\frac{25/4}{4}} = 2\sqrt{\frac{25}{16}} = 2 	imes \frac{5}{4} = \frac{5}{2} = 2.5$.

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

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