有條直線族既問題

The family of straight line of L1 and L2:
2x + y - 1 + k (3x-2y-5) = 0 (where k is a +ve constant) ...(0)

The intersection point of L3 and L4:

x - 3 = 0...(1)
x + y -1 = 0...(2)

By solving them, x = 3 and y = -2

So, L5 passes through the point (3, -2)

Subsitude (3, -2) into (0)
6 - 2 -1 +k (8) = 0
3 + 8k = 0
So, k = -3/8

Re-subsitude k = -3/8 into (0)
2x + y - 1 + (-3/8) (3x - 2y - 5) = 0
2x +y - 1 - 9/8 x + 6/8 y + 15/8 = 0
16x + 8y - 8 - 9x + 6y + 15 = 0
7x + 14y + 7 = 0
x + 2y + 1 = 0

The equation of straight line of L5 is x + 2y + 1 = 0