Find the equations of lines parallel to `3x-4y-5=0` at a unit distance from it.

Explanation

Equation of any line parallel to
`3x-4y-5=0` is `3x-4y-lamda=0`
Let us find a point on the given line.
Let `x=-1,` then `y=-2`
Thus `(-1,2)` is a point on `3x-4y-5=0`
Since the distance between two lines is 1 unit.
`therefore` The length of the perpendicular from `(-1,-2)` to `3x-4y+lamda=0` is 1.
`implies(|3(-1)-4(-2)+lamda|)/(sqrt(3^(2)+(-4)^(2)))=1`
`implies(|5+lamda|)/(5)=1`
`implies|5+lamda|=5`
`implies5+lamda=+-5`
`implieslamda=0orlamda=-10`
Thus, the equations of the required lines are
`3x-4y=0and3x-4y-10=0.`