How to find the vertices of a trinagle given points on the triangle?

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Emmanuel 2014-5-28

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评论： Emmanuel 2014-5-29

I have been given 6 points where 2 points define a side of the trianglee. how can I obtain the vertices of triangle?

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Matt J 2014-5-28

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Meaning the 6 points consist of duplicates of the vertices? If so, this looks applicable

http://www.mathworks.com/matlabcentral/fileexchange/8354-consolidator

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Matt J 2014-5-28

编辑：Matt J 2014-5-28

Are the red points located at the vertices or just at arbitrary locations on the edges? Do you know which pairs of red points belong to the same edge of the triangle?

Emmanuel 2014-5-28

yes! red points ar enot locate don the vertices , but on the sides of the triangle.Every side of the triangle will pass through the red points. One side has two red points, so d whole triangle will have six red points. Using these red points, I have to create the lines, and then form a triangle and find the vertices.

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Matt J 2014-5-28

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编辑：Matt J 2014-5-28

在 MATLAB Online 中打开

If S is a 6x2 matrix such that each pair of consecutive rows S(i,:) and S(i+1,:) are points on one side of the triangle, then you can use qlcon2vert ( Available Here ) as follows,

     x0=mean(S).'; %interior point
     A=[S(2,:)-S(1,:); S(4,:)-S(3,:); S(6,:)-S(5,:)]*[0 -1; 1 0];  
     b=sum(A.*S(1:2:end,:),2);
     D=diag(sign(b-A*x0));
     A=D*A; b=D*b;
    Vertices=qlcon2vert(x0,A,b)

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Matt J 2014-5-28

I cannot read it. To use the

button properly, highlight the code first and then click the button.

Emmanuel 2014-5-29

在 MATLAB Online 中打开

This is the code which I have written. Ignore the mathematical error. I am not able to plot the triangle and obtain the vertices

 for i = 1 : 3
  [xp] =   input('enter the value of 1st co-ord');
  [yp] = input('enter the value of 2nd co=ord');
  %for a and b:
  yp(1) = xp(2); xp(2) = yp(1);
  a = (yp(2)-yp(1)) / (xp(2)-xp(1)); 
  b = yp(1)-a*xp(1); 
  %y = a*x + b; 
  xlims = [0 15]; 
  ylims = [0 15];
  y = xlims*a+b; 
  z = line( xlims, y );
  xlim(xlims); ylim(ylims);
  end