如图：A(aX, aY)，B(bX, bY)，BC=L，求C点坐标(x, y)
思路：
(1)
1-1 kAB * kCB = -1 // 互相垂直两直线斜率的乘积 = -1
1-2 [(aY - bY) / (aX - bX)] * [(y - bY) / (x - bX)] = -1
1-3 (aY - bY) * (y – bY) = -(aX – bX) * (x - bX)
1-4 (y – bY) = - [(aX - bX) * (x - bX)] / (aY - bY)
1-5 y = bY - [(aX - bX) * (x - bX)] / (aY - bY)
(2)
2-1 CB = L
2-2 CB² = L²
2-3 (x – bX)² + (y – bY)² = L²
(3)
把1-4带入2-3
3-1 (x – bX)² + {-[(aX – bX) * (x – bX)] / (aY – bY)}² = L²
3-2 (x – bX)² + [(aX – bX)² * (x – bX)²] / (aY – bY)² = L²
3-3 (x – bX)² + [1 + (aX – bX)² / (aY – bY)²] = L²
3-4 (x – bX)² + {[aY - bY]² + (aX – bX)² * (aY - bY)²} / (aY – bY)²} = L²
3-5 (x - bX)² = [L² * (aY – bY)²] / [(aX – bX)² + (aY – bY)²]
3-6 x – bX = ±√{[L² * (aY - bY)²] / [(aX - bX)² + (aY – bY)²]}
3-7 x – bX = ±[L * (aY - bY)] / √[(aX - bX)² + (aY – bY)²]
3-8 x = bX ± [L * (aY - bY)] / √[(aX - bX)² + (aY – bY)²]
(4)
把3-8带入1-5
4-1 y = bY – {(aX - bX) * (bX ± [L * (aY – bY)] / √[(aX - bX)² + (aY – bY)²] – bX)} / (aY – bY)
4-2 y = bY – {(aX - bX) * (±[L * (aY – bY)] / √[(aX - bX)² + (aY – bY)²])} / (aY – bY)
4-3 y = bY ± [L * (aX - bX)] / √[(aX - bX)² + (aY – bY)²]
结果：
x1 = bX + [L * (aY - bY)] / √[(aX - bX)² + (aY – bY)²]
y1 = bY - [L * (aX - bX)] / √[(aX - bX)² + (aY – bY)²]
和
x1 = bX - [L * (aY - bY)] / √[(aX - bX)² + (aY – bY)²]
y1 = bY + [L * (aX - bX)] / √[(aX - bX)² + (aY – bY)²]