The perpendicular distance between the lines | vedclass

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(A) The given equation is $(x-2y+1)^2 + k(x-2y+1) = 0$. Factoring this,we get $(x-2y+1)(x-2y+1+k) = 0$. This represents two parallel lines: $L_1: x-2y

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The given equation is $(x-2y+1)^2 + k(x-2y+1) = 0$. Factoring this,we get $(x-2y+1)(x-2y+1+k) = 0$. This represents two parallel lines: $L_1: x-2y+1 = 0$ $L_2: x-2y+(1+k) = 0$. 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,$A=1, B=-2, C_1=1, C_2=1+k$. Given $d = \sqrt{5}$,so $\sqrt{5} = \frac{|1-(1+k)|}{\sqrt{1^2+(-2)^2}}$. $\sqrt{5} = \frac{|-k|}{\sqrt{5}}$. $|-k| = \sqrt{5} 	imes \sqrt{5} = 5$. $|k| = 5$,so $k = \pm 5$. Since the options only provide $5$,the correct option is $A$.

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

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