Question

The distance between the parallel lines given by $(x+7y{)}^2+4\sqrt{2}(x+7y)-42=0$ is

• A. $\frac{4}{5}$
• B. $4\sqrt{2}$
• C. $2$
• D. $10\sqrt{2}$

Answer 1

Answer: (C)

$2$

Let the two parallel lines be

$ax+by+c=0$

and $ax+by+c_1=0$

$\therefore$ Equation of pair of straight lines will be

$(ax+by+c)(ax+by+c_1)=0$

$\Rightarrow {(ax+by)}^2+(c+c_1)(ax+by)+c{c}_1=0$

On comparing with given equation

$c_1+c=4\sqrt{2}$

and $c{c}_1=-42$

$\therefore c_1-c=\sqrt{{\left(4\sqrt{2}\right)}^2-4(-42)}$

$=10\sqrt{2}$

Now, distance between them $=\left|\frac{c_1-c}{\sqrt{a^2+b^2}}\right|$

$=\frac{10\sqrt{2}}{5\sqrt{2}}$

$=2$