Consider the two ladders two lines on a graph. Let the shorter ladder extend from
(0,0) to (6,8). Let the longer ladder extend from (6,0) to (0,108^{1/2}).
The square root of 108 can be found using the pythagorean formula.
Then solve for the slope and y intercept to find the equations of the two lines:

The shorter ladder have the equation y=(4/3)*x.

The longer ladder has the equation y=(-108^{1/2}/6)*x + 108^{1/2}.

The lines meet where the ladders cross.

Use substitution to solve for y.

**Acknowledgement:** Thanks to Steve Lutz for suggesting this
method of solution.

Scott R. Walshon pointed out that if x and y are the heights where the ladders touch the wall, and p is the height of the intersection then 1/p = 1/x + 1/y.

To prove this let x be the height on the right wall and y the height on the left wall. Call a the distance from the left edge of the alley to the point on the alley directly below the intersection point of the ladders. Call b the distance from the right edge of the alley to the point on the alley directly below the intersection point of the ladders. From similar triangles we get:

p/x = a/(a+b)

p/y = b/(a+b)

Add the two equations:

p/x + p/y = a/(a+b) + b/(a+b)

py/xy + px/xy = (a+b)/(a+b)

p(x+y)/xy = 1

p = xy/x+y

1/p = x+y/xy

1/p = x/xy + y/xy

1/p = 1/y + 1/x

Thanks Scott for this observation.

Michael Shackleford, A.S.A.