Question

The distance between the 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

The correct option is C $2$

Given equation of pair of straight lines is

$(x+7y{)}^2+4\sqrt{2}(x+7y)-42=0$

$\Rightarrow x^2+49y^2+14xy+4\sqrt{2}x+28\sqrt{2}y-42=0$

Here $a=1,b=49,h=7,g=2\sqrt{2},f=14\sqrt{2},c=-42$
Here $h^2=ab$

The given equation represents a pair of parallel lines.

$d=2\sqrt{\frac{g^2-ac}{a(a+b)}}$

$\Rightarrow d=2$