Question

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. $\frac{\sqrt{5}}{2}$
• E. $\sqrt{\frac{5}{2}}$

Answer 1

Answer: (A)

$\because$ The given equation of pair of parallel lines is $x^2+4xy+4y^2+3x+6y-4=0$
Here, $a=1,b=4g=\frac{3}{2}$ and $c=-4.$
Then, required distance $=2\sqrt{\frac{g^2-ac}{a(a+b)}}$
$=2\sqrt{\frac{\frac{9}{4}+4}{5}}=\frac{2.\sqrt{16+9}}{2.\sqrt{5}}$
$=\frac{2\sqrt{25}}{2\sqrt{5}}=\frac{5}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}$
$=\sqrt{5}$