Q. The distance between the pair of parallel lines $x^{2} + 4xy + 4y^{2} + 3x + 6y - 4 = 0$ is
A) $\sqrt{5}$
B) $\frac{2}{\sqrt{5}}$
C) $\frac{1}{\sqrt{5}}$
D) $\frac{\sqrt{5}}{2}$
Explanation
(a)
$Here,\ a = 1,\ b = 4,\ g = \frac{3}{2},\ f = 3,\ h = 2\ and\ c = - 4$
$Then,\ required\ distance = 2\sqrt{\frac{\frac{9}{4} + 4}{5}}$
$= \frac{2\sqrt{25}}{2\sqrt{5}} = \frac{5}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \sqrt{5}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.