What is the distance between the lines: x/2 + y + 2 = 0 and 4x + 8y = -13

It is possible to find the distance between two lines only
if they are parallel to each other.

The lines we have are:
x/2 + y + 2 = 0 and 4x + 8y = -13.

Convert them to the
slope-intercept form y = mx + c

x/2 + y + 2 =
0

=> y = -x/2 - 2

4x +
8y = -13

=> 8y = -4x -
13

=> y = -x/2 -
13/8

The slope of both the lines is -1/2, they are
parallel.

The product of the slope of perpendicular lines
is equal to -1. Take a line perpendicular to the two given lines like y =
2x

Find the point of intersection of y = 2x with y = -x/2 -
2

2x = -x/2 - 2

=> 5x/2
= -2

=> x = -4/5

y =
-8/5

The point of intersection is (-4/5,
-8/5)

Find the point of intersection of y=2x and y = -x/2 -
13/8

2x = -x/2 -
13/8

=> 5x/2 =
-13/8

=> x = -13/20

y =
-13/10

The  point of intersection is (-13/20,
-13/10)

The distance between (-4/5, -8/5) and (-13/20,
-13/10) is

sqrt[(-13/20 + 4/5)^2 + (-13/10 +
8/5)^2]

=>
0.3354

The distance between the two lines is
0.3354