Find the distance between the pair of parallel lines `x^2+4x y+4y^2+3x+6y-4=0`

Explanation

Given lines are
`(x+2y)^(2)+3(x+2y)-4=0`
`:. x+2y=(-3+-sqrt(9+16))/(2)=(-3+-5)/(2)=-4,1`
therefore , the lines are
`x+2y+4=0`
`x+2y-1=0`
Required distance `=(|4-(-1)|)/(sqrt(1+4))`
`=(5)/(sqrt(5))=sqrt(5)`