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. $\frac{2}{5}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\pm 3\sqrt{5}$
• D. None of these

Answer 1

The correct option is C

$\pm 3\sqrt{5}$

Find the value of $k$:

Given,

The distance between the lines $3$.

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

$\Rightarrow \left(x-2y\right)\left(x-2y+k\right)=0$

$x-2y=0$ and $x-2y+k=0$

These are two parallel lines.

Distance between parallel lines$=\frac{\left|c_1-c_2\right|}{\sqrt{a^2+b^2}}=3$

$$\begin{gathered} \Rightarrow \frac{\left|k-0\right|}{\sqrt{1^2+{\left(-2\right)}^2}}=3 \\ \Rightarrow k=\pm 3\sqrt{5} \end{gathered}$$

Hence, the correct option is (C).