How to define random points above and below a define line within a specific range

1. Suppose line equation is given by y=mx+c and i have a random set of values of x
2. Then we can get corresponding values of y
3. Now with the line defined I would like to define random points above and below the line and within a specific range

[Roger Stafford]

@Santosh Prakash: You haven’t stated in what way the statistical density of points above your line is to differ from that below the line. In other words, what role does your line play in the placement of your random points? Also you need to explain in greater detail your phrase "within a specific range."

[Santosh Prakash]

I would like to have the points uniformly distributed about the line. The line defines what angle or orientation I need the points. More on the application I intend to use. I need to model blood vessel with points representing the components randomly placed within the vessel. And the orientation of the blood vessel can be variable. The line is to be used as the center of the vessel with particles around it. Range of points or particles is defined by the radius of vessel.

[Santosh Prakash]

Attached required rough design.

Line is the dashed line. I require particles uniformly distributed about the line(represented by the rectangular region) Then I need to check the particles if they are within the vessel(represented with the cylinder) and define a velocity model based on the distance from the line.

Answer 1

[Image Analyst]

Do you just want to add uniformly distributed noise, like

y = y + amplitude * rand(1, length(y));

[Santosh Prakash]

I am sorry this is not what I wanted to know. I have explained more in detail now to avoid confusion. Thanks

[Image Analyst]

I read your next post and I think I did describe what you want. Try this:

numPoints = 50;
x = sort(5 * rand(1, numPoints), 'ascend');
m = 0.4;
c = 3;
y = m*x+c;
plot(x, y, 'bd-', 'LineWidth', 2);
grid on;
hold on;
amplitude = 1.6; % Whatever...
noise = amplitude * rand(1, length(y)) - amplitude / 2;
plot(x, noise, 'ms-'); % Plot noise
yNoisy = y + noise;
plot(x, yNoisy, 'r.', 'LineWidth', 2, 'MarkerSize', 10);
ax = gca;
ax.XAxisLocation = 'origin';

Tell me what's wrong with it. How does it differ from what you said?

[Santosh Prakash]

Thank you @Image Analyst. This is what I was looking for.

Just to be on safer side I will check with my adviser once and I will accept the answer tomorrow.

[Roger Stafford]

@Image Analyst. I would guess that Santosh would prefer the points be located in a perpendicular, not vertical, direction from points on the given line segment. As it stands, your code will produce a parallelogram shape of points rather than a rectangular shape. Try it with 10000 points and you will see.

[Image Analyst]

I had considered that but since he didn't say so I just went with the easiest way. Of course you could just multiply the noise by the rotation matrix to "tilt" it perpendicular to the noise. Since it's random, I don't know that it will make much difference to have vertical noise, unless you want multiple noise points for each x values - in that case you'd notice a difference. Like if each x point had 30 noise points instead of only 1. Again that wasn't specified so I just assumed 1 noise point per x point, but it's easy to generalize.

It would have been so helpful if Santosh had just posted a diagram.

[Santosh Prakash]

Posted a diagram for reference

[Image Analyst]

I think the code, as-is, should do the job. I don't think tilting the random locations would lead to a substantially different random arrangement.

Answer 2

[Roger Stafford]

Let w be the allowed perpendicular distance of random points on either side of the line segment. Let x1 be the x-coordinate of left end of segment and d be the segment’s length. Let N be the desired number of random points.

   a = atan(m);
   x = d*cos(a)*rand(N,1)+x1;
   y = m*x+c;
   h = 2*w*rand(N,1)-w;
   x = x - h*sin(a);
   y = y + h*cos(a);
   X = [x1,x1+d*cos(a)];
   Y = m*X+c;
   plot(x,y,'y.',X,Y,'r-',X,Y,'r*')
   axis equal

[Santosh Prakash]

Thank you @Roger

My professor feels the code suggested by Image Analyst is sufficient.