Find the distance between y=-x and y=-x-4.

-- On a graph, those equations produce two lines. 

-- The lines are parallel because they have the same slope ( -1 ). 

-- The y-intercept of one line is zero, so it goes through the origin.

-- The y-intercept of the other line is -4 , so it crosses the y-axis at 4  units
below the origin.

-- The two lines cross the y-axis at two points that are  4  units apart vertically,
so the two lines are  4 vertical units apart everywhere.

-- If you go through the same kind of argument, you discover that they're
also  4  horizontal units apart everywhere.  I think this happens because
their slopes are ' 1 ', (alright, ' -1 '), and you would find that their x-intercepts
are also ' 1 ' and ' -1 '.

-- But if you want to get technical about it . . .

You simply asked for 'the distance'.  I gave you the horizontal and vertical
distances between two sloping parallel lines. A good math teacher would throw
this out.  The 'distance' between parallel lines is defined as the perpendicular
distance ... the length of a line that's perpendicular to both of them.

The first method that pops into my mind goes like this:

-- The slope of these two parallel lines is  -1 , so the slope of a line that's
perpendicular to them is  +1 .

-- Let's look at the line that's perpendicular to them and passes through
the origin.  Its equation is  Y = X , and it hits one of the parallel lines there
at the origin.

-- Where does it hit the other parallel line ?  Well, that line is  y = -x-4 ,
and we want the point where  y=x  intersects it.  So  x = -x - 4 .
Add 'x' to each side:  2x = -4 .  Divide each side by 2:  x = -2 .
The point is  (-2, -2) .

-- What's the length of the line between the two parallel lines that's
perpendicular to them ?  We know 2 points on it . . . (0, 0) and (-2,-2) .
The distance between those points is . . .

     √ [ (-2-0)² + (-2-0)² ] = √(4+4) = √8  =  ( √4 ) ( √2 ) =  2 √2 .

That's about  2.8284... , and that's the real distance between those
two parallel lines.