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

Correct option is C

Given:

​$$x^2 - 6xy + 9y^2 + 3x - 9y - 4 = 0 \\[6pt]$$​​

Formula used :

​$$d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$$​​

Solution:

​$$\text{Group quadratic terms:} \\x^2 - 6xy + 9y^2 = (x - 3y)^2 \\[6pt]\text{So the equation becomes:} \\\ \\(x - 3y)^2 + 3x - 9y - 4 = 0 \\= (x - 3y)^2 + 3(x - 3y) - 4 = 0 \\[6pt]\text{Let } z = x - 3y \Rightarrow z^2 + 3z - 4 = 0 \\\ \\z = \frac{-3 \pm \sqrt{3^2 + 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{-3 \pm \sqrt{25}}{2} = \frac{-3 \pm 5}{2} \\[6pt]\Rightarrow z_1 = 1,\quad z_2 = -4 \Rightarrow x - 3y = 1,\ x - 3y = -4 \\[6pt]\text{Now apply the distance formula:} \\\ \\d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} \quad \text{where } a = 1,\ b = -3,\ c_1 = -1,\ c_2 = 4 \\\ \\d = \frac{| -1 - 4 |}{\sqrt{1^2 + (-3)^2}} = \frac{5}{\sqrt{10}} = \frac{5\sqrt{10}}{10} = \frac{\sqrt{10}}{2}=\sqrt{\frac 52} \\[10pt]$$

d = ​$$\sqrt{\frac 52}$$​