Distance between point and a line (from two points)

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 1

Try using the norm function from numpy.linalg

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

Method 2

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.

Method 3

For the above-mentioned answers to work, the points need to be numpy arrays, here’s a working example:

import numpy as np
p1=np.array([0,0])
p2=np.array([10,10])
p3=np.array([5,7])
d=np.cross(p2-p1,p3-p1)/np.linalg.norm(p2-p1)

Method 4

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

Method 5

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.

Method 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

Method 7

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.

Method 8

Cross products are helpful for the 2D case, but they do not generalize well to other dimensions. Dot products do however. The dot product of two orthogonal vectors is zero in any space, which you can use to come up with a simple solution.

Let’s say you have P4 on the same line as P1–P2. You could parametrize it with parameter t such that

P4 = P1 + t * (P2 - P1)

The goal is to find P4 such that

(P3 - P4) . (P2 - P1) == 0

Expanding P4 in terms of t and simplifying:

(P3 - P1 - t * (P2 - P1)) . (P2 - P1) == 0
(P3 - P1) . (P2 - P1) == t * ||P2 - P1||^2
t = (P3 - P1) . (P2 - P1) / ||P2 - P1||^2

You therefore have

D = ||P3 - P4|| = ||P3 - (P3 - P1) . (P2 - P1) / (||P2 - P1||^2)||

I’ve written a function in my library of utility routines called haggis. You can use haggis.math.segment_distance to compute the distance to the entire line (not just the bounded line segment) like this:

d = haggis.math.segment_distance(P3, P1, P2, segment=False)

Method 9

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

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

0 0 votes

Article Rating