If the line is defined by the implicit equation ax+by=c (i.e. ax+by-c=0) and the the normal vector (a,b) is normalised (i.e. a2+b2=1) then the signed distance of a point x,y from the line is just d=ax+by-c.

If the distances from A,B,C,D are dA,dB,dC,dD, then you have

-a*hl + b*hl - c = dA (1)
a*hl + b*hl - c = dB (2)
a*hl - b*hl - c = dC (3)
-a*hl - b*hl - c = dD (4)

Subtracting 1 from 2

2*a*hl = dB - dA
=> a = (dB - dA) / (2*hl)

Subtracting 3 from 2

2*b*hl = dB - dC
=> b = (dB - dC) / (2*hl)

Adding 1 and 3

-2*c = dA + dC
=> c = -(dA + dC) / 2

Alternatively, adding 2 and 4 gives the same result:

-2*c = dD + dB
=> c = -(dB + dC) / 2

I.e. one of the distances is redundant; the mean of the distances for opposite corners must be equal (and is equal to the distance of the line from the origin).

Note that if a2+b2 isn’t equal to 1, you still have the correct equation for the line; it just means that your signed “distances” include a scale factor.