Python 在给定斜率的线上找到给定距离处的点（Find points at a given distance on a line of given slope）

给定二维点 p(x 0 , y 0 )的坐标。找到距离该点 L 的点，使得连接这些点所形成的线的斜率为M。

例子： 
输入： p = (2, 1)
        L = sqrt(2)
        M = 1
输出：3, 2
        1, 0
解释：
与源的距离为 sqrt(2) ，并具有所需的斜率m = 1。

输入： p = (1, 0)
        L = 5
        M = 0
输出： 6, 0
        -4, 0
        
我们需要找到与给定点距离为 L 的两个点，它们位于斜率为 M 的直线上。
这个想法已在下面的帖子中介绍:
C++ https://blog.youkuaiyun.com/hefeng_aspnet/article/details/141320964

Java https://blog.youkuaiyun.com/hefeng_aspnet/article/details/141321133

Python https://blog.youkuaiyun.com/hefeng_aspnet/article/details/141321178

C# https://blog.youkuaiyun.com/hefeng_aspnet/article/details/141321206

Javascript https://blog.youkuaiyun.com/hefeng_aspnet/article/details/141321238

根据输入的斜率，该问题可以分为 3 类。  

1、如果斜率为零，我们只需要调整源点的 x 坐标

2、如果斜率无限大，则需要调整 y 坐标

3、对于其他斜率值，我们可以使用以下方程来找到点

现在利用上述公式我们可以找到所需的点。

# Python program to find the points on a line of
# slope M at distance L
import math
 
 
# structure to represent a co-ordinate
# point
 
 
class Point:
    def __init__(self, x, y):
        self.x = x
        self.y = y
 
# Function to print pair of points at
# distance 'l' and having a slope 'm'
# from the source
 
 
def printPoints(source, l, m):
    # m is the slope of line, and the
    # required Point lies distance l
    # away from the source Point
    a = Point(0, 0)
    b = Point(0, 0)
 
    # slope is 0
    if m == 0:
        a.x = source.x + l
        a.y = source.y
 
        b.x = source.x - l
        b.y = source.y
 
    # if slope is infinite
    elif math.isfinite(m) is False:
        a.x = source.x
        a.y = source.y + l
 
        b.x = source.x
        b.y = source.y - l
    else:
        dx = (l / math.sqrt(1 + (m * m)))
        dy = m * dx
        a.x = source.x + dx
        a.y = source.y + dy
        b.x = source.x - dx
        b.y = source.y - dy
 
    # print the first Point
    print(f"{a.x}, {a.y}")
 
    # print the second Point
    print(f"{b.x}, {b.y}")
 
 
# driver function
p = Point(2, 1)
q = Point(1, 0)
printPoints(p, math.sqrt(2), 1)
print("\n")
printPoints(q, 5, 0)
 
# The code is contributed by Gautam goel(gautamgoel962)

输出：
3, 2 
1, 0 

6, 0 
-4, 0

时间复杂度： O(1)

辅助空间： O(1)