Find the value(s) of k such that the distance | vedclass

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(B) Given equation is $(x-2y)^2 + k(x-2y) = 0$.Factoring the equation,we get $(x-2y)(x-2y+k) = 0$.This represents two parallel lines: $L_1: x-2y = 0$

Vedclass · Accepted Answer

$\pm 3\sqrt{5}$

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(B) Given equation is $(x-2y)^2 + k(x-2y) = 0$.Factoring the equation,we get $(x-2y)(x-2y+k) = 0$.This represents two parallel lines: $L_1: x-2y = 0$ 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,$A=1, B=-2, C_1=0, C_2=k$.So,$3 = \frac{|k-0|}{\sqrt{1^2+(-2)^2}} = \frac{|k|}{\sqrt{5}}$.Therefore,$|k| = 3\sqrt{5}$,which implies $k = \pm 3\sqrt{5}$.

Find the value$(s)$ of $k$ such that the distance between the two parallel lines represented by $(x-2y)^2 + k(x-2y) = 0$ is $3$ units.

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