Draw a triangle in image knowing its center of gravity

i need to draw a white triangle in a black image. So the first idea that i had got is to define the three peaks that define the triangle and then recuperate the points that belong to its surface. and then affect the value 1 to these points in a black image. To recuperate the points belong to the surface of the triangle i write this code which reflects the idea that a point belongs to a plan it verifies an equation system like after: ( the plane will be the triangle)

if true
  k=1;
for i=1:size_image_x
  for j=1:size_image_y

     equation_x=i-x_peak1==xk*(x_peak2-x_peak1)+yk*(x_peak3-x_peak1);
     equation_y=j-y_peak1==xk*(y_peak2-y_peak1)+yk*(y_peak3-y_peak1);
     eqx=solve(equation_x,xk,yk);
     eqy=solve(equation_y,xk,yk);

    if isempty(eqx)~=1 && isempty(eqy)~=1
      point_x(1,k)=i;
      point_y(1,k)=j;
      k=k+1;
     end

    end
    end
  end
but it doesn't work
may any one help me please 
thanks

Answer 1

[Image Analyst]

Way too complicated. Simply use poly2mask() and do it all in one line of code:

mask = poly2mask([x_peak1, x_peak2, x_peak3], [y_peak1, y_peak2, y_peak3], size_image_y, size_image_x);

[jan]

okay thanks a lot it is a very good and simple solution you are right. do you have any other solution if we don't have the three peaks of the triangle but we have its center of gravity????

[Image Analyst]

You need something else, like the radius and angle to one of the vertices or something. Obviously you could just spin a triangle around the center of gravity and they would all be triangles, just at different orientations. It's just a matter of simple trigonometry to get the three vertices once you have enough other information. What orientation do you want to draw it at? Surely you must have SOME idea.