A line isn't made up of pixels. The pixels in its trajectory are modified taking line width and antialiasing into account. Drawing a line with default settings and zooming in on the image looks like the image below. Very few pixels get the full 100% of the given color. Lots of pixels are changed.
![zoomed in image of drawn line]()
Depending on your final goal, you could calculate pixel coordinates using the method described in the post you linked (note that the pixels on a saved image can deviate a bit from the pixels on-screen). And then use e.g. Bresenham's line algorithm to find the coordinates of points in-between. Note that a naive Bresenham's algorithm would draw a 45 degree line much thinner looking than a horizontal line. On a modern screen a one-pixel wide line would be almost invisible.
Here is a possible Bresenham-like interpretation of the linked code:
import numpy as np
import matplotlib.pyplot as plt
def points_in_line(x0, y0, x1, y1):
dx = np.round(np.abs(x1 - x0))
dy = np.round(np.abs(y1 - y0))
steps = int(np.round(max(dx, dy))) + 1
return np.vstack([np.linspace(x0, x1, steps), np.linspace(y0, y1, steps)]).T
fig, ax = plt.subplots()
points, = ax.plot([0, 1, 2, 4, 5, 6, 9], [0, 5, 3, 2, 2, 9, 8], 'b-')
ax.axis([-1, 10, -1, 10])
# Get the x and y data and transform them into pixel coordinates
x, y = points.get_data()
xy_pixels = ax.transData.transform(np.vstack([x, y]).T)
x_pix, y_pix = xy_pixels.T
# find all points in each line
all_pix = [points_in_line(x0, y0, x1, y1) for x0, y0, x1, y1 in zip(x_pix[:-1], y_pix[:-1], x_pix[1:], y_pix[1:])]
all_x_pix, all_y_pix = np.concatenate(all_pix).T
# In matplotlib, 0,0 is the lower left corner, whereas it's usually the upper
# left for most image software, so we'll flip the y-coords...
width, height = fig.canvas.get_width_height()
all_y_pix = height - all_y_pix
print('Coordinates of the lines in pixel coordinates...')
for xp, yp in zip(all_x_pix, all_y_pix):
print(f'{x:0.2f}\t{y:0.2f}')
# save the figure with its current DPI
fig.savefig('test.png', dpi=fig.dpi)