幾何坐標的問題

(a) x + 3y - 2 + k(x - y - 6) = 0
x + 3y - 2 + kx - ky - 6k = 0
(1+k)x + (3-k)y - (2+ 6k) = 0
(k-3)y = (k+1)x - (2+ 6k)
y = [(k+1)/(k-3)]x - [(2+6k)/(k-3)]

所以, L 的斜率是 (k+1)/(k-3)

(b) x + 2y - 4 = 0
2y = - x + 4
y = (-1/2)x + 2
x + 2y - 4 = 0 的斜率是 -1/2

L 平行於 x + 2y - 4 = 0, 所以
L 的斜率 = -1/2
(k+1)/(k-3) = -1/2
2(k+1) = - (k-3)
2k + 2 = -k + 3
3k = 1
k = 1/3

(c) x + 3y - 2 + k(x - y - 6) = 0
(1+k)x + (3-k)y - (2+ 6k) = 0 ............(1)

設 k = -1, (1) 成為:
[1+(-1)]x + [3-(-1)]y - [2+ 6(-1)] = 0
4y +4 = 0
y = -1

設 k = 3, (1) 成為:
(1+3)x + [3-(3)]y - [2+ 6(3)] = 0
4x -20 = 0
x = 5

所以 P 的坐標 是 (5 , -1)