The distance between the pair of parallel lin | vedclass

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(D) The given equation is $x^2 + 2xy + y^2 - 8ax - 8ay - 9a^2 = 0$. This can be written as $(x+y)^2 - 8a(x+y) - 9a^2 = 0$. Let $X = x+y$,then $X^2 - 8

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$5\sqrt{2}a$

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(D) The given equation is $x^2 + 2xy + y^2 - 8ax - 8ay - 9a^2 = 0$. This can be written as $(x+y)^2 - 8a(x+y) - 9a^2 = 0$. Let $X = x+y$,then $X^2 - 8aX - 9a^2 = 0$. Factoring the quadratic,we get $(X - 9a)(X + a) = 0$. So,the two lines are $x + y - 9a = 0$ and $x + y + a = 0$. 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,$A = 1, B = 1, C_1 = -9a, C_2 = a$. $d = \frac{|-9a - a|}{\sqrt{1^2 + 1^2}} = \frac{|-10a|}{\sqrt{2}} = \frac{10a}{\sqrt{2}} = 5\sqrt{2}a$.

The distance between the pair of parallel lines $x^2 + 2xy + y^2 - 8ax - 8ay - 9a^2 = 0$ is

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