Question

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

• A. $5\sqrt{5}$
• B. $3$
• C. $3\sqrt{5}$
• D. $-3\sqrt{5}$

Answer 1

Answer: (C)

$3\sqrt{5}$

Given eq

$(x-2y{)}^2+k(x-2y)=0$

$distance=3$

Now, distance between these two lines would be ;

$d=|\frac{k-0}{\sqrt{1+4}}|=\frac{|k|}{\sqrt{5}}$

or, $d=3\Rightarrow |k|=3\sqrt{5}$