How can I get the perpendicular distances between each boundary point to the hough line?

Question

This is my input picture.

The yellow points on the curve are given by the "drawcountours" method and the blue straight line is obtained by the hough line. here is the following code:

import numpy as np
import cv2
from matplotlib import pyplot as plt
# from scipy.cluster.vq import vq, kmeans
import numpy as np
import glob



images = glob.glob(r'C:\Users\Desktop\dist.png')

for fname in images:

  Image = cv2.imread(fname)

  gray = cv2.cvtColor(Image, cv2.COLOR_BGR2GRAY)
  ret,seg = cv2.threshold(gray,40,250,cv2.THRESH_BINARY)

  # cv2.imshow('H',seg)
  # cv2.waitKey(0)


  total=0
  _, contours, hierarchy = cv2.findContours(seg, cv2.RETR_TREE, cv2.CHAIN_APPROX_SIMPLE)
  for c in contours:
    if cv2.contourArea(c) >80 and cv2.contourArea(c) <2000:
      M = cv2.moments(c)
      cX = int(M["m10"] / M["m00"])
      cY = int(M["m01"] / M["m00"])

      epsilon = 0.001 * cv2.arcLength(c, True)
      approx = cv2.approxPolyDP(c, epsilon, True)
      total+=1

      cv2.drawContours(Image, approx, -1, (0, 255, 255), 5)

  cv2.imshow('l',Image)
  cv2.waitKey(0)

  lines = cv2.HoughLines(seg, 7, np.pi / 270, 300)

  for rho, theta in lines[0]:
    a = np.cos(theta)
    b = np.sin(theta)
    x0 = a * rho
    y0 = b * rho
    x1 = int(x0 + 1000* (-b))
    y1 = int(y0 + 1000* (a))
    x2 = int(x0 -1000 * (-b))
    y2 = int(y0 -1000 * (a))

    cv2.line(Image, (x1, y1), (x2,y2), (255, 255, 0), 2)
  cv2.imshow('l', Image)
  cv2.waitKey(0)

 ######get line equation#####
  points = [(x1, y1), (x2, y2)]
  x_coords, y_coords = zip(*points)
  A = vstack([x_coords, ones(len(x_coords))]).T
  m, c = lstsq(A, y_coords)[0]
  print("Line Solution is y = {m}x + {c}".format(m=m, c=c))
 
  #####to convert the obtained line equation to Ax+By+C=0 form###### 
  A = m
  B = -1
  C = 0
  for i in range(1, approx.shape[0]):
      
      
      per_dis = ((A * approx[i][0][0]) + (B * approx[i][0][1]) + C) / math.sqrt((A ** 2) + (B ** 2)) # the perpendicular distance
      inter_x = int(((B * (approx[i][0][0] - (A * approx[i][0][1]))) - (A * C)) / ((A ** 2) + (B ** 2)))#the x coordinate of the intersection point
      inter_y = int((A * ((-B * approx[i][0][0]) + (A * approx[i][0][1])) - (B * C)) / ((A ** 2) + (B ** 2)))#the y coordinate of the intersection point
     
      cv2.line(Image, (approx[i][0][0],approx[i][0][1]), (inter_x,inter_y), (255, 0, 255), 2)# to draw the green line 

  cv2.imshow('l', Image)
  cv2.waitKey(0)

How can I obtain the perpendicular distances from those contour points( yellow points) to the hough line? In other words, I want the following output.

I want to display the green lines and also want to know the length of each green line.

I tried calculating the the perpendicular distance using the formulae from Wikipedia but got the following output:

Answer 1

In order to get that result, you will first need to understand a few concepts. The first concept is the line vector representation. You need to first find the vector representation of your line.

First you need to get 2 random points, a1 and a2, on your Hough Line to create a vector representation, A,

A = a2 - a1

and vector B from

B = b1 - a1

With these vector you can calculate the magnitude and angle,θ using the following formula:

Magnitude = (A^T ⋅ B) / |A|

θ = cos^-1((A^T ⋅ B)/ |A||B|)

Then, using the trigonometry function, you should be able to find the distance L, which is the distance between the point to the Hough Line.