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(B) The given equation is $9x^2 - 6xy + y^2 + 18x - 6y + 8 = 0$.This can be written as $(3x - y)^2 + 6(3x - y) + 8 = 0$.Let $3x - y = t$,then the equa

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$2/\sqrt{10}$

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(B) The given equation is $9x^2 - 6xy + y^2 + 18x - 6y + 8 = 0$.This can be written as $(3x - y)^2 + 6(3x - y) + 8 = 0$.Let $3x - y = t$,then the equation becomes $t^2 + 6t + 8 = 0$.Factoring the quadratic equation: $(t + 4)(t + 2) = 0$.So,the two parallel lines are $3x - y + 4 = 0$ and $3x - y + 2 = 0$.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,$A = 3$,$B = -1$,$C_1 = 4$,and $C_2 = 2$.$d = \frac{|4 - 2|}{\sqrt{3^2 + (-1)^2}} = \frac{2}{\sqrt{9 + 1}} = \frac{2}{\sqrt{10}}$.

The distance between the parallel lines represented by the equation $9x^2 - 6xy + y^2 + 18x - 6y + 8 = 0$ is

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