Question

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

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is C

Applying the formula, the distance between them is $\left|2\sqrt{\frac{\left(\frac{k^2}{4}\right)-0}{1.(1+4)}}\right|=|\frac{k}{\sqrt{5}}|$,

$\therefore |\frac{k}{\sqrt{5}}|=3\Rightarrow k=\pm 3\sqrt{5}$