The distance between the lines `4x + 3y = 11` and `8x + 6y = 15` is

Explanation

`4x+3y-11=0`
`8x+6y-15=0`
`implies4x+3y-15/2=0`
The required distance `=(|(-15)/(2)-(-11)|)/(sqrt(4^(2)+3^(2)))`
`=(|(-15)/(2)-(-11)|)/(5)`
`=(7)/(2xx5)=7/10`units