题解 | 点到直线距离

import math

class Point:
    def __init__(self, x, y):
        self.x = x
        self.y = y

class Line:
    def __init__(self, point_a, point_b):
        self.point_a = point_a
        self.point_b = point_b

def get_distance(P, L):
    # TODO: 计算点P到直线L的距离
    # pass

##方法1，向量 h = 丨ap丨 * sin& ,sin& = 丨ap × ab丨/ (丨ap丨 * 丨ab丨) , h = 丨ap × ab丨/丨ab丨
## ab = (L.point_b.x-L.point_a.x ,L.point_b.y - L.point_a.y)
##ap = (p.x -L.point_a.x, p.y - L.point_a.y )
#----------------------------------------------------------------------
    x_ab ,y_ab = (L.point_b.x-L.point_a.x) ,(L.point_b.y - L.point_a.y)
    x_ap , y_ap = (P.x -L.point_a.x ) , (P.y - L.point_a.y )
    chaji = x_ab * y_ap - x_ap * y_ab
    m_ab = math.sqrt(x_ab**2 + y_ab**2)
    h = abs(chaji) / m_ab
    return h



##方法2，点斜式直线 + 点到直线的距离公式=abs(A*x_0 + B*y_0 + C)/ (sqrt(A**2 + B**2))
#----------------------------------------------------------------------
    if L.point_b.x == L.point_a.x :
        h = abs(P.x - L.point_a.x)
    else:
        k = (L.point_b.y - L.point_a.y) / (L.point_b.x - L.point_a.x)
        c = L.point_a.y - k * L.point_a.x
        # 直线L的点斜式公式为：kx-y+C = 0
        h = abs(k * P.x - P.y + c) / math.sqrt(k**2 + 1)
    return h



def main():
    a, b = map(int, input().split())
    sx, sy, tx, ty = map(int, input().split())
    
    point_a = Point(sx, sy)
    point_b = Point(tx, ty)
    point_c = Point(a, b)
    line = Line(point_a, point_b)
    
    print("{:.2f}".format(get_distance(point_c, line)))

if __name__ == "__main__":
    main()