Find the intersection between object boundaries and a line

I have a binary mask and a line that passes through the mask as shown in the attached figure. I want to find the intersection point between the mask and the line (the red plus point in the figure).

Can you please help me with this?

I have a binary mask and a line that passes through the mask as shown in the attached figure. I want to find the intersection point between the mask and the line (the red plus point in the figure).

Can you please help me with this?

This is the code to generate the output image:

    mask = imread(‘mask.png’);
    bw = bwareafilt(im2bw(mask));
    s = regionprops(bw, 'Centroid', 'Extrema');
    corners = s.Extrema;
    Pt1dx = (corners(4,1) + corners(5,1))/2;
    Pt1dy = (corners(4,2) + corners(5,2))/2;
    Pt2dx = s.Centroid(1);
    Pt2dy = s.Centroid(2);
    figure;
    imshow(bw);
    hold on;
    plot(Pt1dx, Pt1dy, 'g+'); 
     
    slope = (Pt1dy-Pt2dy)/(Pt1dx-Pt2dx);
    Targetdx = -75.590551181 * cos(slope) + Pt1dx;
    Targetdy = -75.590551181 * sin(slope) + Pt1dy;
    hold on;
    plot(Targetdx, Targetdy, 'b*');
    hold on;
    line([Pt1dx, Pt1dy], [Targetdx, Targetdy])
    hold on;
     
     
    slopeInv = -1/slope; 
    xLine = 1:size(bw,2);
    yLine = slopeInv.*(xLine - Targetdx) + Targetdy;  
    plot(xLine, yLine, 'b');   %# Plot the perpendicular line
     

[Ghada Alzamzmi]

The output image.

[Image Analyst]

Yes, but what about the input image mask.png. You forgot to attach that.

Answer 1

[Ghada Alzamzmi]

I was able to find the points using the script below:

https://www.mathworks.com/matlabcentral/fileexchange/11837-fast-and-robust-curve-intersections

Thank you all for your responses.

Answer 2

[Matt J]

B=cell2mat(bwboundaries(bw,8,'noholes'));

[x,y]=deal(B(:,2),size(bw,1)+1-B(:,1));

[d,loc]=min( abs(  slopeInv.*(x - Targetdx) + Targetdy -y)  );

intersection  = [x(loc), y(loc)]

[Ghada Alzamzmi]

Thanks a lot Matt for your quick response!

I followed your steps and I got the point of the other side. I want the second intersection point.

Please see the attached image.

This is the code:

    
    B = cell2mat(bwboundaries(bw,8,'noholes'));
    [x,y]=deal(B(:,2),size(bw,1)+1-B(:,1));
    [d,loc]=min( abs(  slopeInv.*(x - Targetdx) + Targetdy -y)  );
    intersection1  = [x(loc), y(loc)];
    hold on
    plot(intersection1(1), intersection1(2), 'g+'); 

[Matt J]

I don't know what general criterion you use to define the "correct side". This version looks for the intersection with the largest y-coordinate:

  
    B = cell2mat(bwboundaries(bw,8,'noholes'));
    [x,y]=deal(B(:,2),size(bw,1)+1-B(:,1));
    

    [a,b,c]=deal(slopeInv,-1, Targetdy-slopeInv.*Targetdx);
    
    loc = abs( (a*x+b*y+c)/norm([a,b]) )  <= sqrt(2)*1.0001;
    [~,ymax]=max( y(loc) );
    
    intersection  = [x(loc(ymax)), y(loc(ymax))];

    hold on
    plot(intersection1(1), intersection1(2), 'g+');