Two lines are given by x^2 - 4xy + 4y^2 + kx - 2ky = 0 , then the value of k so that the distance between them is 3 is

Step-by-step solution

The given equation is x^2 - 4xy + 4y^2 + kx - 2ky = 0 This can be rewritten as (x - 2y)^2 + k(x - 2y) = 0 (x - 2y)(x - 2y + k) = 0 The two lines are x - 2y = 0 and x - 2y + k = 0 The distance between these parallel lines is given by d = |c₁ - c₂| a^2 + b^2 d = |k - 0| 1^2 + (-2)^2 = |k| 5 Given that the distance is 3 , we have: |k| 5 = 3 |k| = 3 5 Answer: 3 5