Find the distance between the parallel lines `3x+4y-7=0` and `3x+4y+8=0`.

Explanation

Put `y=0` in `3x+4y-7=0`
`3x+0-7=0`
`implies x=(7)/(3)`
`:.` Point `((7)/(3),0)` lies on this line
Now, the length of perpendicular from `((7)/(3),0)` to the second line `3x+4y+8=0` is
`=|(3*(7)/(3)+0+8)/(sqrt(3^(2)+4^(2)))|=|(15)/(5)|=3` units.