Have you graphed this? You really need to graph this to see what is going on. Sorry I cannot draw it for an answer.
Notice they intersect in the 1st quadrant.
For L1 to cross the pos x-axis and the neg y-axis means it has a positive slope. In slope intercept form, this would mean that for y=mx+b, the m MUST BE POSITIVE.
For L2 to cross the neg x-axis and the pos y-axis means it also has a positive slope.
If you rearange each linear equation to be in the form of y = mx + b, you get
L1: y = (-1/a)x - (b/a) So the slope is (-1/a) which must be positive, so a must be negative.
L2: y = (-1/c)x - (d/c) So the slope is (-1/c) which must be positive, so c must be negative.
The final term of y=mx+b, the b is the y-intercept.
L1 crosses the negative y-axis, so -(b/a) must be negative. a is negative, so be must also be negative so that -(b/a) is negative.
L2 crosses the positive y-axis, so -(d/c) must be positive, but we already know that c is negative, so d must be positive.
As for a - c: I graphed it and used real numbers and found that a - c = positive.
a = neg
b = neg
c = neg
d = pos
a-c = pos
A graph is REALLY HELPFUL for this problem!