F.5 直線方程

1.
代 y = 0 入 L_1 得 x = 6, 即 A = (6, 0)
代 x = 0 入 L_1 得 y = 4, 即 B = (0, 4)
L_1 的斜率為 -2/3 即 L_2 的斜率是 3/2
故 L_2 的方程為 y = (3/2)x
代入 L_1 得 2x + 3(3/2)x - 12 = 0
x = 24/13, y = 36/13
即 C = (24/13, 36/13)
現求通過 O, B, C 三點的圓。
假設方程為
x² + y² + Dx + Ey + F = 0
代入 O(0, 0), B(0, 4), C(24/13, 36/13) 得
{ F = 0
{ 16 + 4E + F = 0
{ (24/13)² + (36/13)² + D(24/13) + E(36/13) + F = 0

式二即 4E = -16 即 E = -4
式三即
(24/13)² + (36/13)² + D(24/13) - 4(36/13) = 0
24² + 36² + D(24)(13) - 4(36)(13) = 0
1872 + 312D - 1872 = 0
312D = 0
D = 0

所求方程是：x² + y² - 4y = 0


2.
(a)
由於比例關係，
OP : PQ = 5 : (b - 5)
〔你畫圖就會明白。〕
〔5 和 b 分別是兩線的 y-軸截距。〕

(b)
L_1 的斜率是 m，故 L_3 的斜率是 -1/m
L_3 的方程是 y = (-1/m)x
代入 L_1：
(-1/m)x = mx + 5
(m + 1/m)x = -5
(m² + 1)x = -5m
x = -5m/(m² + 1)
y = (-1/m)[-5m/(m² + 1)] = 5/(m² + 1)
P = ( -5m/(m² + 1) , 5/(m² + 1) )

(c)
先計算 OP
= √{ [-5m/(m² + 1)]² + [5/(m² + 1)]² }
= √{ [-5m]² + [5]² } / (m² + 1)
= 5 √(m² + 1²) / (m² + 1)
= 5/√(m² + 1²)

由於 OP : PQ = 5 : (b - 5)
即 OP / PQ = 5 / (b - 5)
所以 PQ = OP(b - 5)/5 = (b - 5)/√(m² + 1²)

2015-03-29 01:08:16 補充：
PQ = (b - 5)/√(m² + 1)

For Q1. For line L1, when x = 0, y = 4, so B is (0,4).
Let C be (x,y).
Slope of OC = y/x
Slope of BC = (y - 4)/(x - 0) = (y - 4)/x
Since OC is perpendicular to BC,
slope OC x slope BC = - 1
(y/x)[(y - 4)/x] = -1
y(y - 4) = - x^2
x^2 + y^2 - 4y = 0 is the equation of the circle.