对于每对相邻顶点 A，B：
构造一个从 A 到 B 的向量，称它为 p
现在构造一个从 A 到您的测试点 X 的向量，称之为 q
一对向量的点积公式是 pq = |p||q|cosC 其中 C 是向量之间的角度。
所以如果 pq/|p||q| == 1 那么点 AX 和 AB 是共线的。在计算机上工作，您将需要 1 - pq/|p||q| < some_small_value 取决于您想要的准确度。
还需要检查 |q| < |p| （即X比B更接近A）
如果 4&5 为真，则您的观点在边界上。
编辑
我认为我已经看到这样做的另一种方法是获取您的测试点 X，并通过 X 构建一条垂直于 A 和 B 之间的线的线。找到这条线和线 A->B 交叉的位置。计算出从 X 到该交叉点的距离，如果足够小，您认为该点在线上。
编辑——有趣的小练习！
由于我记错了一些数学，早先发布了一些错误的代码。在回家的火车上玩 Pythonista 并想出了这个似乎基本上可行的方法。因为在 iPad 上编辑帖子很痛苦，所以留下了数学证明！
没有做太多的测试，没有测试除以零等，警告用户。
# we determine the point of intersection X between
# the line between A and B and a line through T
# that is perpendicular to the line AB (can't draw perpendicular
# in ascii, you'll have to imagine that angle between AB and XT is 90
# degrees.
#
# B
# /
#. X
# / \
# / T
# A
# once we know X we can work out the closest the line AB
# comes to T, if that distance is 0 (or small enough)
# we can consider T to be on the line
import math
# work out where the line through test point t
# that is perpendicular to ab crosses ab
#
# inputs must be 2-tuples or 2-element lists of floats (x,y)
# returns (x,y) of point of intersection
def intersection_of_perpendicular(a,b,t):
if a[0] == b[0]:
return (a[0],t[1])
if a[1] == b[1]:
return (t[0],a[1])
m = (a[1] - b[1])/(a[0] - b[0]) #slope of ab
x_inter = (t[1] - a[1] + m*a[0] + (1/m)*t[0])*m/(m**2 + 1)
y_inter = m*(x_inter - a[0]) + a[1]
y_inter2 = -(1/m)*(x_inter - t[0]) + t[1]
#print '...computed ',m,(x_inter, y_inter), y_inter2
return (x_inter, y_inter)
# basic Pythagorean formula for distance between two points
def distance(a,b):
return math.sqrt( (a[0]-b[0])**2 + (a[1]-b[1])**2 )
# check if a point is within the box defined by a,b at
# diagonally opposite corners
def point_in_box(a,b,t):
xmin = min(a[0],b[0])
xmax = max(a[0],b[0])
ymin = min(a[1],b[1])
ymax = max(a[1],b[1])
x_in_bounds = True
if xmax != xmin:
x_in_bounds = xmin <= t[0] <= xmax
y_in_bounds = True
if ymax != ymin:
y_in_bounds = ymin <= t[1] <= ymax
return x_in_bounds and y_in_bounds
# determine if point t is within 'tolerance' distance
# of the line between a and b
# returns Boolean
def is_on_line_between(a,b,t,tolerance=0.01):
intersect = intersection_of_perpendicular(a,b,t)
dist = distance(intersect, t)
in_bounds = point_in_box(a,b,t)
return in_bounds and (dist < tolerance)
a = (0,0)
b = (2,2)
t = (0,2)
p = intersection_of_perpendicular(a,b,t)
bounded = point_in_box(a,b,t)
print 'd ',distance(p,t), ' p ',p, bounded
a = (0,2)
b = (2,2)
t = (1,3)
p = intersection_of_perpendicular(a,b,t)
bounded = point_in_box(a,b,t)
print 'd ',distance(p,t),' p ',p, bounded
a = (0.0,2.0)
b = (2.0,7.0)
t = (1.7,6.5)
p = intersection_of_perpendicular(a,b,t)
bounded = point_in_box(a,b,t)
on = is_on_line_between(a,b,t,0.2)
print 'd ',distance(p,t),' p ',p, bounded,on