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

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

Discussion

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