First I will write L1 and L2 in slope-intercept form:
L1: y = 1/2*x - 3
L2: y = -x + 4
1. L3 is parallel to L1 and has the same y-intercept as L2.
The slope of L1 is m1 = 1/2
The y-intercept of L2 is b = 4
From these, L3 becomes
y = 1/2*x + 4
2. L4 is perpendicular to L1 and has the same x-intercept as L2.
The slope perpendicular to L1 is m2 = -1/m1 = -2
The x-intercept of L2 is found by setting y = 0: 0 = -x + 4 --> x = 4, which gives us the point (4,0)
From these, L4 becomes
y = -2x + b ... plug in (4,0)
0 = -2(4) + b --> b = 8
y = -2x + 8
# The equation of L2 is x = -1. L1 is parallel to L2 and contains the point (3,5).
L2: x = -1
A like parallel to this has the same form as this equation: x = c
Using this form, we can solve for c using the point given: (3,5)
From these, L1 becomes
x = c ... plug in (3,5)
3 = c
x = 3
That is pretty straight forward. It's just a vertical line in the form x = c passing through the point (3,5). The only vertical line that passes through this point is the line x = 3.