The ratio in which the line 3x+4y+2=0 divides the distance between 3x+4y+5=0 and 3x+4y-5=0 is?

Answer

0.12986111111111

Explanation

Lines 3x+4y+2=0 and 3x+4y+5=0 are on the same side of the origin.
The distance between these lines is
`d_(1) = |(2-5)/(sqrt(3^(2)+4^(2)))| = (3)/(5)`
Lines 3x+4y+2=0 and 3x+4y-5=0 are on the opposite side of the origin.
The distance between these lines is
`d_(2) = |(2+5)/(sqrt(3^(2)+4^(2)))| = (7)/(5)`
Thus, 3x+4y+2=0 divides the distance between 3x+4y+5 = 0 and 3x+4y-5=0 in the ratio `d_(1) : d_(2), i.e., 3:7.`