Draw a straight line in image, given an angle

Is there a way to draw a straight line in an image from a starting (x,y) coordinate, such that the line is equivalent to a specified angle?

For example, if I start at pixel (50,50), how can I draw a line to be 30 degrees of this pixel? I would like to draw 10 pixels in this direction (obviously, the resulting line will be zig-zagged)...

[Walter Roberson]

This is just another way of phrasing your previous question, http://www.mathworks.com/matlabcentral/answers/16986-normal-vector-of-line-based-on-pixel-coordinates

[Philip]

Not precisely, as the angle of the line to be drawn varies as opposed to being a static 90 degrees. I need a way of adapting the function to work for any angle, which I couldn't do from the previous post. I found an alternative method using imrotate to rotate the image until the line to be drawn is completely straight, but the interpolation causes too many problems when rotating back to the original orientation.

[Image Analyst]

I fail to see why the answers you got there won't apply to lines at any angle. The fact is, they will apply.

[Walter Roberson]

Irrelevant, Philip. In both cases you identify a specific starting point and need to draw a line at an angle from that point. It matters not to the general algorithm whether the angle has a 90 degree relationship to another known angle, or if it has (say) a 42 degree relationship to the other known angle.

[Philip]

But what happens if I don't know the end points? What if all I know is the start point, angle of the line to be drawn, and the length of the line to be drawn? It seems many of the solutions rely on a start and end point in order to make the line... I obviously misinterpreted the previous answers as I couldn't get it to work for my function and didn't think it could be applied when only the above is known.

[Image Analyst]

You know your line endpoints. You said you knew the (x,y) starting point. And the other endpoint's x position, if you don't know it, can't be less than 1 or more than the number of columns in your image. Therefore you know both endpoints.

[Walter Roberson]

Calculate the total angle relative to the forward Y axis (i.e., angle 0 on a polar plot.) Choose a large radius, R, such as 1000, for an imaginary line segment; don't worry if it is out of the image. Now, R*sin(theta) and R*cos(theta) give you the total delta x and total delta y if you were drawing the whole line segment. Round to integers, and add those rounded deltas to the starting point to get the imaginary end point. Now apply Bresenham's line algorithm to the imaginary line between the starting point and the identified imaginary end point, except stop when you have generated as many total pixels as you desire. The result will be a line segment drawn from the starting point at your desired angle, for the desired number of pixels.

[Philip]

Thanks for your kind suggestions. Mathematics is not my strong point, but your comments have helped me to get the otherwise unknown coordinates. I will now see if I can implement Bresenham's line algorithm to connect the two points!

Much appreciated!

Answer 1

[Image Analyst]

How about this:

lineLength = 10;
angle = 30;
x(1) = 50;
y(1) = 50;
x(2) = x(1) + lineLength * cosd(angle);
y(2) = y(1) + lineLength * sind(angle);
hold on; % Don't blow away the image.
plot(x, y);
xlim([0 70]);
ylim([0 70]);
grid on;

If you want the line to be "burned into" your image, then you'll have to use imline() - search answers for my demo on that - or use the Bresenham line algorithm like Walter suggested if you want to do it yourself.

[Philip]

Thank you for your kind suggestion! This is a great way of obtaining the end points of the line, which I can then use in combination with the start points to draw the line at the relevant pixel positions in between.

One question - can imline be used without requiring user input, or could I even use improfile to do this? Otherwise I will attempt an implementation of the Bresenham line algorithm as per Walter's suggestion.

Thanks again!

[Image Analyst]

Philip: Yes you can. See my demo in an answer below.

[K M Ibrahim Khalilullah]

@Image Analyst I have the same problem. Thanks for your suggestion. >> Is it possible to control end point according to the Image size? because i need end point also.

[Image Analyst]

Yes, just make up your formula's for (x1,y1) and (x2,y2) using the number of rows and columns in your image and whatever else you need to do to set the endpoints wherever you want them.

[Michelle De Luna]

@Image Analyst: Hi there! Quick question (and this may be an easy one, so I apologize in advance, but I would greatly appreciate your guidance): Is there a way to create two lines of identical length when the angles dictating the incline of each of the lines is different? I am in the middle of drawing two equidistant lines on a plot using your suggested code above, but each line starts at a different point and has a different angle. Here's what I have so far. Any suggestions?

lineLength = 10
angle_one = 22.0443
t(1) = 410
e(1) = 200
t(2) = t(1) + lineLength * cosd(angle_one)
e(2) = e(1) + lineLength* sind(angle_one)
plot(t,e)
grid on
hold on

angle_two = 51.2134
c(1) = 436
d(1) = 217
c(2) = c(1) + lineLength * cosd(angle_two)
d(2) = d(1) + lineLength * sind(angle_two)
plot(c,d)
grid on

[Image Analyst]

Yes, of course, because you specified different starting points and angles. What do you want? The same starting point and the same angles? Then they would overlap.

[Michelle De Luna]

Right. The starting points can differ (as should the angles). In my scenario, it would be perfectly fine to have the two lines overlap. I would merely need the lines to have similar length, or to have a length of 21 pixels each.

[Image Analyst]

I don't see the problem. The code is doing exactly what you told it to. Nor do I know what you want. Perhaps a diagram would help. Have it show what you are getting, and what you want instead.

[Walter Roberson]

Michelle De Luna:

No. It is possible to generate two lines of identical length only if they happen to have a particular Pythagorean triple relationship such as listed at http://oeis.org/A084646 or related sequences. For example if the distance is 25 then there are two different angles that can generate that identical distance.

Any case in which the angle and distance does not happen to correspond to one of the triples, you can never get out identical distances.

Answer 2

[Image Analyst]

% Demo to write an ellipse and a line into the overlay of an image,
% and then to burn those overlays into the image.

%----- Initializing steps -----
% Clean up
clc;
clear all;
close all;
workspace; % Display the workspace panel.

% Change the current folder to the folder of this m-file.
if(~isdeployed)
  cd(fileparts(which(mfilename)));
end

hasIPT = license('test', 'image_toolbox');
if ~hasIPT
  % User does not have the toolbox installed.
  message = sprintf('Sorry, but you do not seem to have the Image Processing Toolbox.\nDo you want to try to continue anyway?');
  reply = questdlg(message, 'Toolbox missing', 'Yes', 'No', 'Yes');
  if strcmpi(reply, 'No')
    % User said No, so exit.
    return;
  end
end

% Display images to prepare for the demo.
monochromeImage = imread('pout.tif');
subplot(2, 4, 1);
imshow(monochromeImage);
title('Original Image');
subplot(2, 4, 2);
imshow(monochromeImage);
title('Original Image with ellipse in overlay');
subplot(2, 4, 5);
imshow(monochromeImage);
title('Original Image');
subplot(2, 4, 6);
imshow(monochromeImage);
title('Original Image with line in overlay');
set(gcf, 'units','normalized','outerposition',[0 0 1 1]); % Maximize figure.
set(gcf,'name','Demo by ImageAnalyst','numbertitle','off') 

%----- Burn ellipse into image -----
% Create elliptical mask, h, as an ROI object over the second image.
subplot(2, 4, 2);
hEllipse = imellipse(gca,[10 10 50 150]); % Second argument defines ellipse shape and position.
% Create a binary image ("mask") from the ROI object.
binaryImage = hEllipse.createMask();
% Display the ellipse mask.
subplot(2, 4, 3);
imshow(binaryImage);
title('Binary mask of the ellipse');
% Let's try to add some text.  (Doesn't work)
% hText = text(50, 100, 'Line of Text');
% textMask = hText.createMask();
% binaryImage = binaryImage & textMask;
% imshow(binaryImage);
% Burn ellipse into image by setting it to 255 wherever the mask is true.
monochromeImage(binaryImage) = 255;
% Display the image with the "burned in" ellipse.
subplot(2, 4, 4);
imshow(monochromeImage);
title('New image with ellipse burned into image');

%----- Burn line into image -----
burnedImage = imread('pout.tif');
% Create line mask, h, as an ROI object over the second image in the bottom row.
subplot(2, 4, 6);
hLine = imline(gca,[10 100],[10 100]); % Second argument defines line endpoints.
% Create a binary image ("mask") from the ROI object.
binaryImage2 = hLine.createMask();
% Display the line mask.
subplot(2, 4, 7);
imshow(binaryImage2);
title('Binary mask of the line');
% Burn line into image by setting it to 255 wherever the mask is true.
burnedImage(binaryImage2) = 255;
% Display the image with the "burned in" line.
subplot(2, 4, 8);
imshow(burnedImage);
title('New image with line burned into image');