Two lines are given by (x - 2y)^2 + k(x - 2y) | vedclass

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(C) The given equation is $(x - 2y)^2 + k(x - 2y) = 0$.This can be factored as $(x - 2y)(x - 2y + k) = 0$.Thus,the two lines are $L_1: x - 2y = 0$ and

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$\pm 3\sqrt{5}$

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(C) The given equation is $(x - 2y)^2 + k(x - 2y) = 0$.This can be factored as $(x - 2y)(x - 2y + k) = 0$.Thus,the two lines are $L_1: x - 2y = 0$ and $L_2: x - 2y + k = 0$.The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.Here,$a = 1$,$b = -2$,$c_1 = 0$,and $c_2 = k$.So,$3 = \frac{|0 - k|}{\sqrt{1^2 + (-2)^2}} = \frac{|k|}{\sqrt{1 + 4}} = \frac{|k|}{\sqrt{5}}$.Therefore,$|k| = 3\sqrt{5}$,which implies $k = \pm 3\sqrt{5}$.

Two lines are given by $(x - 2y)^2 + k(x - 2y) = 0$. The value of $k$ so that the distance between them is $3$,is

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