The perpendicular distance between the lines given by (x-2 y +1)^2+ k (x-2 y +1)=0 is 5 , then k =

Step-by-step solution

Let L = x - 2y + 1 so the equation becomes L^2 + kL = 0 . Factoring yields L(L + k) = 0 , producing the parallel lines x - 2y + 1 = 0 and x - 2y + (1 + k) = 0 . The distance between parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0 is d = |C₁ - C₂| A^2 + B^2 . Applying this with A = 1 , B = -2 , C₁ = 1 , and C₂ = 1 + k : 5 = |1 - (1 + k)| 1 + 4 = |-k| 5 Multiplying both sides by 5 gives 5 = |-k| . The absolute value equation implies k = 5 or k = -5 , but since only k = 5 appears among the options, the solution i