f.4 amaths

L1: x - y + 1 = 0
L2: x + 2y + 4 = 0
設直線族方程是:
x - y + 1 + k(x + 2y + 4) = 0
(k+1)x + (2k -1)y + (4k+1) = 0
(1-2k)y = (k+1)x + (4k+1) = 0
y = [(k+1)/(1-2k)]x + [(4k+1)/(1-2k)] .......................(1)
這直線族方程的斜率 = (k+1)/(1 -2k)

族中斜率為3, 所以
(k+1)/(1 -2k) = 3
k+1 = 3(1-2k)
k+1 = 3 - 6k
7k = 2
k = 2/7 ........................(2)

該族中斜率為3的直線方程, 將(2)代入(1)
y = {[(2/7)+1]/[1-2(2/7)]}x + {[4(2/7)+1]/[1-2(2/7]}
y = 3x + 5
3x -y + 5 = 0

所以, 該族中斜率為3的直線方程是　3x -y + 5 = 0