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

1 view (last 30 days)
I have been given 6 points where 2 points define a side of the trianglee. how can I obtain the vertices of triangle?

Answers (2)

Matt J
Matt J on 28 May 2014
Meaning the 6 points consist of duplicates of the vertices? If so, this looks applicable
  5 Comments
Matt J
Matt J on 28 May 2014
Edited: Matt J on 28 May 2014
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
Emmanuel on 28 May 2014
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.

Sign in to comment.


Matt J
Matt J on 28 May 2014
Edited: Matt J on 28 May 2014
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)
  3 Comments
Matt J
Matt J on 28 May 2014
I cannot read it. To use the
button properly, highlight the code first and then click the button.
Emmanuel
Emmanuel on 29 May 2014
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

Sign in to comment.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!