Distance between point and a line (from two points)

Question

I\'m using Python+Numpy (can maybe also use Scipy) and have three 2D points

(P1, P2, P3); 

I am trying to get the distance from P3 perpend

Answer 1

Shortest Distance from Point to a Line

This is the code I got from https://www.geeksforgeeks.org:

import math 

# Function to find distance 
def shortest_distance(x1, y1, a, b, c):    
    d = abs((a * x1 + b * y1 + c)) / (math.sqrt(a * a + b * b)) 
    print("Perpendicular distance is", d)

Now you have to find A, B, C, x, and y.

import numpy as np
closest = []
x = (x ,y)
y = (x, y)
coef = np.polyfit(x, y, 1)
A = coef[0]
B = coef[1]
C = A*x[0] + B*x[1]

Now you can plug in the values:

shortest_dis = shortest_distance(x, y, A, B, C)

The full code may look like this:

import math
import numpy as np

def shortest_distance(x1, y1, a, b, c):    
    d = abs((a * x1 + b * y1 + c)) / (math.sqrt(a * a + b * b)) 
    print("Perpendicular distance is", d)

closest = []
x = (x ,y)
y = (x, y)
coef = np.polyfit(x, y, 1)
A = coef[0]
B = coef[1]
C = A*x[0] + B*x[1]
shortest_dis = shortest_distance(x, y, A, B, C)

Please let me know if any of this is unclear.

Answer 2

Based on the accepted answer

Test with below line equation -

Find the perpendicular distance from the point (5, 6) to the line −2x + 3y + 4 = 0

• x-intercept p1 = [0, -4/3]
• y-intercept p2 = [2, 0]
• shortest distance from p3 = [5, 6] = 3.328

import numpy as np
norm = np.linalg.norm

p1 = np.array([0,-4/3])
p2 = np.array([2, 0])

p3 = np.array([5, 6])
d = np.abs(norm(np.cross(p2-p1, p1-p3)))/norm(p2-p1)
# output d = 3.328201177351375

Answer 3

3D distance should use np.dot def threeD_corres(points_3_d,pre_points_3_d,points_camera):

  for j in  range (0,len(pre_points_3_d)):
      vec1 = list(map(lambda x:x[0]- x[1],zip(pre_points_3_d[j], points_camera)))
      vec2 = list(map(lambda x:x[0]- x[1],zip(pre_points_3_d[j], points_3_d[j])))
      vec3 =  list(map(lambda x:x[0]- x[1],zip(points_3_d[j], points_camera)))
      distance = np.abs(np.dot(vec1_1,vec2_2))/np.linalg.norm(vec3)

      print("#########distance:\n",distance)
  return  distance

Answer 4

abs((x2-x1)*(y1-y0) - (x1-x0)*(y2-y1)) / np.sqrt(np.square(x2-x1) + np.square(y2-y1))

Can be used directly through the formula, just have to plug in the values and boom it will work.

Answer 5

np.cross returns the z-coordinate of the cross product only for 2D vectors. So the first norm in the accepted answer is not needed, and is actually dangerous if p3 is an array of vectors rather than a single vector. Best just to use

d=np.cross(p2-p1,p3-p1)/norm(p2-p1)

which for an array of points p3 will give you an array of distances from the line.

Answer 6

To find distance to line from point if you have slope and intercept you can use formula from wiki https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line Python:

def distance(point,coef):
    return abs((coef[0]*point[0])-point[1]+coef[1])/math.sqrt((coef[0]*coef[0])+1)

coef is a tuple with slope and intercept