It'd have easier for us to solve the problem if you've cared to post the diagram here. The site needs user login to view the diagram.

Anyway, I'm assuming the following diagram.


x and y-intercept of L2 is x and y respectively and h is the shortest distance between the two lines. As the slope of the line is 45 degrees, x and y-intercept of L2 will be equal.

Hence, x = y.

Now we need to find h.
We can redraw the figure as follows,


ABC is a right angled isosceles triangle with angle BAC being the right angle and AC = AB = x.

If we observe carefully, triangle ACD is also a right angled isosceles triangle with angle ADC being the right angle and AD = CD = h
Hence, AC = √(AD² + CD²) = √(h² + h²) = √(2h²) = (√2)h
----> x = (√2)h
----> h = x/√2

Now, replace x with the original x or y-intercept as given in the diagram.

Another approach:

• BC = √(AB² + AC²) = √(x² + x²) = √(2x²) = (√2)x
  
  Now area of the triangle ABC = AB*AC/2 = AD*BC/2
  --> x*x/2 = h*(√2)x/2
  --> x = (√2)h
  --> h = x/√2

A more advanced approach:
Let us assume the x and y-intercept of the line L2 is p. We already know both of them will be same. Hence, the equation of line L2 : y = x + p

The shortest is nothing but the length of the perpendicular drawn on L2 from the origin. Length of the perpendicular drawn from a point (x1, y1) on a line with equation Ax + By + C = 0 is given by the following formula,


In our case the point is (0, 0) and the equation is x - y + p = 0
Hence, the length of the perpendicular = |c|/√(1² + 1²) = |c|/√2

Hence. shortest distance between L1 and L2 is |c|/√2.
Replace c with the original x or y-intercept as given in the diagram.