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Pair (L)^2 + k(L) = 0 at a given distance: find k or the lines

This type of question, asked 3 times: 19 Apr 2025 S2, 11 Apr 2026 S2, 18 Apr 2026 S1. All 3 questions below, oldest first.

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  1. Question 1

    19 Apr 2025 S2

    116
    The perpendicular distance between the lines given by (x−2y+1)2+k(x−2y+1)=0(x- 2y+ 1)^{2}+ k(x- 2y+ 1) = 0 is 5\sqrt{5}, then k=k =

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  2. Question 2

    11 Apr 2026 S2

    57
    If the distance between the lines represented by (x−2y)2+k(x−2y)=0(x - 2y)^2 + k(x - 2y) = 0 is 33 units and k>0k > 0 then the equations of the lines are

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  3. Question 3

    18 Apr 2026 S1

    55
    Two lines are given by x2−4xy+4y2+kx−2ky=0x^2 - 4xy + 4y^2 + kx - 2ky = 0, then the value of k so that the distance between them is 3 is

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