Dear Christophe,
No, changing tolerances won't help. Your problem is ill-posed. Because the ideal triangles only touch at a single point, small perturbations of the input data can make the triangles intersect at multiple points or not at all.
This is why, in the Description section above, I mentioned "It is to be emphasized that A,b must define a solid region". Your intersection point is not solid in R^2, so the inequalities [A1;A2],[b1;b2] are not a legal way of expressing it.

Dear Matt,
tried calculating the intersection between a two triangles touching each other in a point but got following error.
Assume can be solved by setting tolerances correct?
Thanks
Christophe
>> [A1,b1,Aeq1,beq1]=vert2lcon([2,0;0,2;0,0]);
>> [A2,b2,Aeq2,beq2]=vert2lcon([1,1;2,1;1,2]);
>> V = lcon2vert([A1;A2],[b1;b2],[],[])
Something's wrong. We should have found a recession vector (bb<0).

@Bhimavarapu,
The code doesn't generate any plots, but there are other utilities on the FEX that you could use for plotting polyhedra, e.g.,
<http://www.mathworks.com/matlabcentral/fileexchange/9261-plot-2d-3d-region>
As for "not getting the vertices", if you mean you are getting empty output V=[], it means your polyhedron appears empty to the code. Hard to say more without seeing what you're doing.

Cong,
Points on the boundary of the polytope are expected to violate the inequalities by small amounts due to finite precision arithmetic. When I compute the violations for your A,b data with the code below, I find that they are indeed very small O(1e-13),
>> V=lcon2vert(A,b);
>> slacks=bsxfun(@minus, b,A*V');
>> violations = slacks(slacks<0)
violations =
1.0e-13 *
-0.1066
-0.0355
-0.5684
-0.0711
-0.0355
-0.0355
-0.5684
-0.0355
-0.2842
-0.5684
-0.1066
-0.2487
-0.3375

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