How is line 1 related to line L?

we let y=-Lx+4, where L > 0
so x = (-1/4)y + 4/L, and 4/L is the x-intercept
L is positive so 4/L can't be -1.
4/L approaches 0 as L approaches infinity, but L must be a finite number.
Therefore c) 6 is a possiblity.

I think the best approach to this problem is to graph it out using a coordinate plane.

If you put a "point" on the y-intercept at 4, and you know that the line "L" is negative, it has to be choice III.

Remember, a negative slope is "going down the stairs" and a positive slope is "going up the stairs" in a coordinate plane. The x-intercept of 6 is the logical choice.

If it was x-intercept was -1, the slope would be positive & there would be an "undefined" slope if the x-coordinate was 0.

Sound good?

Th egeneral eq of line is y=mx + c at y intercept x=0 so y=c=4
now the slope m is given to be negative so let us assume -1
so now to find x intercept we put y=0 in the eq
ie 0 = -x + 4 we get x as 4 so x is positive
Also we should note that x cannot be zero since it will mean that the line passes through origin and with y intercept given as 4 that is not possible
hence 3