如何使用 Python 在给定斜率和截距坐标 (x, y) 的图像上绘制一条线

Question

I have a slope of a line, and I have the x and y coordinates of the intercept that I want the line to go through. Essentially, I want the line to be displayed on the image itself.

The following is basically the only code I have so far (just the variables):

slope = 3
intercepts = [7, 10]

I want to use Python for this task. Any help would be greatly appreciated!

Answer 1

You can custom this answer from Tommaso Di Noto :

import matplotlib.pyplot as plt
import random

x_intercept = 7  
y_intercept = 10 
my_slope = 3 

def find_second_point(slope,x0,y0):
    # this function returns two points which belongs to the line that has the slope 
    # inserted by the user and that intercepts the point (x0,y0) inserted by the user
    q = y0 - (slope*x0)  # calculate q
    new_x1 = x0 + random.randint(x0,x0+10)  
    new_y1 = (slope*new_x1) + q  
    new_x2 = x0 - random.randint(x0,x0+10)  
    new_y2 = (slope*new_x2) + q 
    return new_x1, new_y1, new_x2, new_y2   


new_x1, new_y1,new_x2, new_y2 = find_second_point(my_slope , x_intercept, y_intercept )

def slope(x1, y1, x2, y2):
    return (y2-y1)/(x2-x1)
print(slope(x_intercept,y_intercept, new_x1, new_y1))


plt.figure(1)  # create new figure
plt.plot((new_x2, new_x1),(new_y2, new_y1), c='r', label='Segment')
plt.scatter(x_intercept, y_intercept, c='b', linewidths=3, label='Intercept')
plt.scatter(new_x1, new_y1, c='g', linewidths=3, label='New Point 1')
plt.scatter(new_x2, new_y2, c='cyan', linewidths=3, label='New Point 2')
plt.legend()  # add legend to image

plt.show()

Output:

slope
3.0