the value of k

Dear Student,

The joint equation of two lines is, {(x - 2 y)}^{2} + k (x - 2 y) = 0 \Rightarrow (x - 2 y) (x - 2 y + k) = 0 \Rightarrow x - 2 y = 0 or x - 2 y + k = 0 So, seperate equation of the lines are : x - 2 y = 0 and x - 2 y + k = 0 Comparing the above equations with a_{1} x + b_{1} y + c_{1} = 0 and a_{2} x + b_{2} y + c_{2} = 0, we get a_{1} = 1; b_{1} = - 2; c_{1} = 0 a_{2} = 1; b_{2} = - 2; c_{2} = k Since, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = 1, so the given lines are parallel . Distance between parallel lines ax + by + c_{1} = 0 and ax + by + c_{2} = 0 is, d = \frac{| c_{1} - c_{2} |}{\sqrt{a^{2} + b^{2}}} Now, c_{1} = 0; c_{2} = k; a = 1; b = - 2; d = 3 So, 3 = \frac{|0 - k|}{\sqrt{{(1)}^{2} + {(- 2)}^{2}}} \Rightarrow k = 3 \sqrt{5}

So, option (B) is correct.

Regards