Q. Given a family of lines $a(2x + y + 4) + b(x - 2y - 3) = 0$, the number of lines belonging to the family at a distance $\sqrt{10}$ from $P(2,\ - 3)$ is
A) 0
B) 1
C) 2
D) 4
Explanation
(b)
$P = \left| \frac{a(4 - 3 + 4) + b(2 + 6 - 3)}{\sqrt{(2a + b)^{2} + (a - 2b)^{2}}} \right| = \sqrt{10}$
$\Rightarrow 25(a + b)^{2} = 10\left( {5a}^{2} + {5b}^{2} \right)$
$\Rightarrow 25(a - b)^{2} = 0 \Rightarrow a = b$
Only one line which is $3x - y + 1 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.