Q85 / 160 · AP EAMCET 2021 (25 Aug Shift 1) · Mathematics
Find the value(s) of $k$ such that the distance between the two parallel lines represented by $(x-2 y)^2+k(x-2 y)=0$ is 3 units
- A0
- Correct answerB$\pm 3 \sqrt{5}$
- C$\pm 5$
- D$\pm 3$
Solution
Given equation,
$$
\begin{aligned}
(x-2 y)^2+k(x-2 y) & =0 \\
\Rightarrow \quad(x-2 y)(x-2 y+k) & =0 \\
(x-2 y)=0 \text { or } x-2 y+k & =0
\end{aligned}
$$
Distance between these parallel lines is
$$
\begin{array}{rlrl}
& \left|\frac{k-0}{\sqrt{1+2^2}}\right| & =\left|\frac{k}{\sqrt{5}}\right|=3 \\
\Rightarrow & k \mid & =3 \sqrt{5} \\
k & = \pm 3 \sqrt{5}
\end{array}
$$
Hence, the correct answer is option: 'B'.