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(C) The given equation is $x^2 - 6xy + 9y^2 + 3x - 9y - 4 = 0$.Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we ge

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$\sqrt{\frac{5}{2}}$

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(C) The given equation is $x^2 - 6xy + 9y^2 + 3x - 9y - 4 = 0$.Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a = 1, h = -3, b = 9, g = 3/2, f = -9/2, c = -4$.The distance between two parallel lines represented by the equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is given by $d = 2\sqrt{\frac{g^2 - ac}{a(a + b)}}$.Substituting the values,we get $d = 2\sqrt{\frac{(3/2)^2 - (1)(-4)}{1(1 + 9)}} = 2\sqrt{\frac{9/4 + 4}{10}} = 2\sqrt{\frac{25/4}{10}} = 2\sqrt{\frac{25}{40}} = 2\sqrt{\frac{5}{8}} = 2 	imes \frac{\sqrt{5}}{2\sqrt{2}} = \sqrt{\frac{5}{2}}$.

The distance between the pair of lines represented by the equation $x^2 - 6xy + 9y^2 + 3x - 9y - 4 = 0$ is

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