Two lines are given by the equation x^2 - 4xy | vedclass

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(A) The given equation is $x^2 - 4xy + 4y^2 + kx - 2ky = 0$.This can be written as $(x - 2y)^2 + k(x - 2y) = 0$.Factoring out $(x - 2y)$, we get $(x -

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$3\sqrt{5}$

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(A) The given equation is $x^2 - 4xy + 4y^2 + kx - 2ky = 0$.This can be written as $(x - 2y)^2 + k(x - 2y) = 0$.Factoring out $(x - 2y)$, we get $(x - 2y)(x - 2y + k) = 0$.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 $c_2 = k$.Given $d = 3$, we have $3 = \frac{|0 - k|}{\sqrt{1^2 + (-2)^2}}$.$3 = \frac{|k|}{\sqrt{1 + 4}} = \frac{|k|}{\sqrt{5}}$.$|k| = 3\sqrt{5}$.Therefore, $k = \pm 3\sqrt{5}$.

Two lines are given by the equation $x^2 - 4xy + 4y^2 + kx - 2ky = 0$. Find the value of $k$ such that the distance between them is $3$.

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