The line L given by x/5 + y/b = 1 passes through the point (13, 32). If the line K is parallel to L and has the equation x/c + y/3 = 1, then the distance between L and K is?

Since line L passes through (13, 32), we have:
13/5 + 32/b = 1
32/b = 1 - 13/5 = -8/5
b = 32 * (-5/8) = -20
The equation of line L is x/5 - y/20 = 1, which can be written as 4x - y = 20.
The slope of line L is 4.
Since line K is parallel to line L, the slope of line K is also 4.
The equation of line K is x/c + y/3 = 1, which can be written as 3x + cy = 3c.
The slope of line K is -3/c = 4, so c = -3/4.
The equation of line K is 3x - (3/4)y = -9/4, which can be written as 12x - 3y = -9.
The distance between two parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 is given by |C1 - C2| / √(A² + B²).
The equations of lines L and K are 4x - y - 20 = 0 and 12x - 3y + 9 = 0, which can be written as 4x - y - 20 = 0 and 4x - y + 3 = 0.
The distance between L and K is |-20 - 3| / √(4² + (-1)²) = 23 / √17 = 23√17 / 17.