如何确定质心点是否接触到线？

如果您的直线具有起点和终点[x1，y1]和[x2，y2]，则直线方程为：

y-y1 = m *（x-x1），其中m =（y2- y1）/（x2-x1）`

然后可以检查一个点是否属于直线，用x或y代替，然后检查另一个是否与直线方程式匹配。

在比顿：

# the two points that define the line
p1 = [1, 6]
p2 = [3, 2]

# extract x's and y's, just for an easy code reading
x1, y1 = p1
x2, y2 = p2

m = (y2-y1)/(x2-x1)

# your centroid
centroid = [2,4]
x3, y3 = centroid

# check if centroid belongs to the line
if (m * (x3-x1) + y1) == y3:
    print("Centroid belongs to line")

但是可能

...计算红色点与线之间的距离（从点到线的距离），然后检查距离是否足够近（即距离小于某个值），将会获得更好的结果。

在Python中：

# points that define the line
p1 = [1, 6]
p2 = [3, 2]
x1, y1 = p1
x2, y2 = p2

centroid = [2,4]
x3, y3 = centroid

# distance from centroid to line
import math # to calculate square root
dist = abs((y2-y1)\*3 - (x2-x1)\*3 + x2\*1 - y2\*1)/math.sqrt((y2-y1)\*2 + (x2-x1)\*2)

if dist < some_value:
    print("Near enough")

回答来源：stackoverflow