Draw line on image with known slope

Given Point A and B and a line equation y = mx + c. I calculated the intercept and slope.

Now I want to draw a line (linspace function) from point A to the image boundary. It worked for lines with small gradients but failed for a nearly 'vertical line'.

I used the following code:

imshow(I)
xlims = xlim(gca);
ylims = ylim(gca);

The problem with vertical lines is shown in the screenshot. What I want is that the blue line stops at the image boundary because there are only 5 magenta 'plus' coordinates within the image, which is too less.

Answer 1

[Adam Danz]

If you have the staring point (A,B), the slope (m) the y-intercept (yint), and the axis limits XL, YL (where XL and YL are (min,max) pairs), you can calculate the end point of the line where it crosses the border of your axes. Then use the start coordinate and end coordinate to draw a line via plot().

% record your axis limits
XL = xlim(gca);
YL = ylim(gca);
hold(gca, 'on')

% calculate the two possible end points
% 1) line crosses upper axis limit. X = ?, Y = YL(2)
xEnd = (YL(2)-yint)/m;             
% 2) line crosses rightward axis limit.  X = XL(2), y = ?
yEnd = (m * XL(2)) + yint ;         

% Now determine which end point to use (ie, which axis it crosses)
if xEnd <= XL(2)
    % The line crosses the upper y axis border
    yEnd = YL(2); 
else
    % The line crosses the rightward x axis border
    xEnd = XL(2); 
end

%Draw the line
plot([A, xEnd], [B, yEnd], 'b-', 'linewidth', 3)

%circle end point for confirmation
plot(xEnd, yEnd, 'mo', 'linewidth', 3)

[Kamu]

Thanks for help, but your solution will just find one intercept between line and image border (lower part) but not the intercept with the upper border. I have attached a screenshot. Ideally, I want to the know the coordinates when the line (here in red) hits the image border.

This is a toy example. I changed YL(2) to YL(1) to get the upper part from the red circle to the upper image border:

img = uint8(zeros(382, 382));
 
pointA = [181.4594, 129.7092];
pointB = [185.3411, 251.6005];
 
imshow(img)
hold on
plot(pointA(1), pointA(2), 'ro')
plot(pointB(1), pointB(2), 'bo')
hold off
 
% From y = mx + x
slope = 31.4015;
intercept = -5.5684e+03;
 
imshow(img)
XL = xlim(gca);
YL = ylim(gca);
hold(gca, 'on')
% calculate the two possible end points
% 1) line crosses upper axis limit. X = ?, Y = YL(2)
xEnd = (YL(1)-intercept)/slope;             
% 2) line crosses rightward axis limit.  X = XL(2), y = ?
yEnd = (slope * XL(1)) + intercept ;         
% Now determine which end point to use (ie, which axis it crosses)
if xEnd <= XL(1)
    % The line crosses the upper y axis border
    yEnd = YL(1); 
else
    % The line crosses the rightward x axis border
    xEnd = XL(1); 
end
%Draw the line
plot([pointA(1), xEnd], [pointA(2), yEnd], 'r-', 'linewidth', 3)
plot([pointB(1), xEnd], [pointB(2), yEnd], 'g-', 'linewidth', 3)
 
X_values = linspace(pointB(1), xEnd);
Y_values = linspace(pointB(2), yEnd);
 
% Check how many coordinates are within image frame
for zz=1:length(X_values)
    hold on
    plot(X_values(zz),Y_values(zz), 'm+');
end

[Adam Danz]

That's because you're using imshow() which flips the y axis. In the image below, note that the max(ylim) is at the bottom of the plot instead of the top.

figure
I = imread('cameraman.tif');
imshow(I)
axis on

To adjust my code in my answer so that the line searches for the intersection with the min(ylim) axis, change the two YL(2) to YL(1) and my solution will work with flipped y axes.

[Kamu]

You are right, that's the solution. Thanks!

[Adam Danz]

Nice, glad it worked!