如何找到一个点的坐标，使其与给定的线成 135 度角？

Question

Given a line segment uv I want to find the coordinates of point x such that they make up an angle of 135 degrees with a given distance of L , below is an illustration:

Answer 1

I would use a triangle to determine where x is, you can use a 45 45 90 triangle because the angle is 135 degrees. If it was not 135 degrees you can use sin and cos to find x

Answer 2

Find uv vector as

uv.x = v.x - u.x
uv.y = v.y - u.y

Calculate its length

luv = sqrt(uv.x * uv.x + uv.y * uv.y)

Find point at continuation of uv line with distance L from v :

px = v.x + L * uv.x  / luv
py = v.y + L * uv.y  / luv

Rotate this point around v by 45 degrees ( Pi/4 ):

qx = v.x + (p.x - v.x) * cos(Pi/4) - (p.y - v.y) * sin(Pi/4)
qy = v.y + (p.x - v.x) * sin(Pi/4) + (p.y - v.y) * cos(Pi/4)

Note that we can simplify tha last expressions a bit using coordinate difference

rx = L * uv.x  / luv
ry = L * uv.y  / luv

Rotate this point around v by 45 degrees ( Pi/4 ):

qx = v.x + rx * cos(Pi/4) - ry * sin(Pi/4)
qy = v.y + rx * sin(Pi/4) + ry * cos(Pi/4)

Also both cos(Pi/4) and sin(Pi/4) are equal to sqrt(2)/2 , so you can use this constant if you don't need different angles.

Answer 3

Actually, you don't have to calculate all the math by yourself. There are several libraries for geometric computation that offer functionality to perform the task in just a couple of lines of code.

For example, Shapely has rotate and scale functions that could be used here:

from shapely.geometry import LineString
from shapely.affinity import rotate, scale

uv = LineString([(0, 0), (1, 0)])
vx_length = 2
pivot_point = uv.boundary[1]
scale_factor = vx_length / uv.length
uv_rotated = rotate(uv, -135, origin=pivot_point)
vx = scale(uv_rotated, 
           xfact=scale_factor,
           yfact=scale_factor,
           origin=pivot_point)
print(vx)
# LINESTRING (2.414213562373095 1.414213562373095, 1 0)
print(vx.length)
# 2.0

Note that we should give the negative angle to rotate since by default rotation is performed counterclokwise but we need it to be clockwise.