解析幾何之對稱問題

以下,尋找對稱點公式免得重複寫N次一樣的步驟(=A=):
設P(x1, y1) 對於 L: Ax + By + C = 0 的對稱點 = P'(x2, y2)
L的斜率 = -A/B
垂直於L的直線L'的斜率 = B/A
L' : (y - y1) = (B/A)(x - x1)
又
P和P'的中點O[(x1 + x2)/2, (y1 + y2)/2] = L和L'的交點

把中點座標代入L, L':
L: A(x1 + x2)/2 + B(y1 + y2)/2 + C = 0
及
L': (y1 + y2)/2 - y1 = (B/A)[(x1 + x2)/2 - x1]
y2 - y1 = (B/A)(x2 - x1)
y2 = (B/A)(x2 - x1) + y1 …(i)

代入L:
A{x1 + x2)/2 + B{y1 + [(B/A)(x2 - x1) + y1]}/2 + C = 0
(A/2 - B^2/2A)x1 + (A/2 + B^2/2A)x2 + By1 + C = 0
[(A^2 - B^2)/2A]x1 + [(A^2 + B^2)/2A]x2 + By1 + C = 0
(A^2 - B^2)x1 + (A^2 + B^2)x2 + 2A(By1 + C) = 0
(A^2 + B^2)x2 = -(A^2 - B^2)x1 -2A(By1 + C)

x2
= [(-A^2 + B^2)x1 -2A(By1 + C)] / (A^2 + B^2)
= {[A^2 + B^2 - 2A^2]x1 - 2A(By1 + C} / (A^2 + B^2)
= x1 - [2A^2x1 + 2A(By1 + C)] / (A^2 + B^2)
= x1 - 2A(Ax1 + By1 + C) / (A^2 + B^2)

將x2 代入 (i):
y2
= (B/A)[x1 - 2A(Ax1 + By1 + C) / (A^2 + B^2) - x1) + y1
= (B/A)[-2A(Ax1 + By1 + C) / (A^2 + B^2)] + y1
= y1 - 2B(Ax1 + By1 + C)/(A^2 + B^2)

即,對稱點P'的座標
= (x2, y2)
= [x1 - 2A(Ax1 + By1 + C) / (A^2 + B^2) , y1 - 2B(Ax1 + By1 + C)/(A^2 + B^2)] ...(ii)

根據(ii),
P(2, 4) 對於 L1 的對稱點Q座標
= {2 - 2(3)[3(2) - 2(4) + 3] / [3^2 + (-2)^2]} , 4 - 2(-2)[3(2) - 2(4) + 3] / [3^2 + (-2)^2]}
= {2 - 6(6 - 8 + 3) / (9 + 4) , 4 + 4(6 - 8 + 3) / (9 + 4)]}
= (20/13, 56/13)
及, P 對於 L2 的對稱點R座標
= {2 - 2(2)[2(2) + 4(4) - 9]/(2^2 + 4^2), 4 - 2(4)[2(2) + 4(4) - 9]/(2^2 + 4^2)}
= {2 - 4(4 + 16 - 9)/(4 + 16), 4 - 8(4 + 16 - 9)/(4 + 16)}
= (-1/5, -2/5)

直線QR方程:
(y - 56/13) / (x - 20/13) = (-2/5 - 56/13) / (-1/5 - 20/13)
(13y - 56) / (13x - 20) = 306/113
1469y - 6328 = 3978x - 6120
306x - 113y + 16 = 0

因為直線QR方程 = 直線AB方程

->求點A的座標:
L1: 3x - 2y + 3 = 0
=> y = (3x + 3)/2
代入y至方程QR:
306x - 113(3x + 3)/2 + 16 = 0
x = 307/273
代入x至L1:
3(307/273) - 2y + 3 = 0
y = 290/91
=> A(307/273, 290/91)

->求點B的座標:
L2: 2x + 4y - 9 = 0
=> y = (9 - 2x)/4
代入y至方程QR:
306x - 113(9 - 2x)/4 + 16 = 0
x = 953/1450
代入x至L2:
2(953/1450) + 4y - 9 = 0
y = 1393/725
=> B(953/1450, 1393/725)
答案: A = (307/273, 290/91), B = (953/1450, 1393/725)