Actually, I am dealing with a problem in barycentric coordinates I got the equations of three lines asenter image description here

I know that these three lines sharing a common point, I know if I prove their det is zero, they are concurrent.

here the problem is to prove det is zero,

my question is there any shorter way to prove that det is zero, or is there any online tool which is useful in proving this, if so please share.

and also how should I find a common point on these lines which is in the form of (x:y:z)

please share the steps to get it or any online tool for finding this.

Thanks in advance.

  • $\begingroup$ Frankly, I'd throw it into Wolfram Alpha. $\endgroup$ – user3482749 Nov 14 '18 at 1:44
  • $\begingroup$ Really then can we find the solution like(x,y,z) from it, if so please share the ans, $\endgroup$ – nimmy Nov 14 '18 at 2:04
  • $\begingroup$ I don't think wolfy like long queries. Use a proper CAS instead $\endgroup$ – user10354138 Nov 14 '18 at 2:05
  • $\begingroup$ O then how to get the solution $\endgroup$ – nimmy Nov 14 '18 at 2:07
  • $\begingroup$ Does $s$ have any relation to $a$, $b$ and $c$? $\endgroup$ – amd Nov 14 '18 at 3:41

Here is the command in Maxima, you can use other CAS if you want:

s: (a+b+c)/2;
M: matrix(

(or use factor instead of ratexpand), which indeed shows the determinant is $0$.

You can also use eigenvalues(M); to see that there is indeed an eigenvalue $0$ of multiplicity $1$. To find the eigenspace, use eigenvectors(M);, which we are only interested in the last part of the output (the full output is [[[eigenvalues],[multiplicities]],[eigenvectors]]) $$ \begin{bmatrix} 1\\ {{bc^3+\left(ab-b^2\right)c^2+\left(-b^3+4ab^2+a^2b \right)c+b^4-ab^3-a^2b^2+a^3b}\over{ac^3+\left(ab-a^2 \right)c^2+\left(ab^2+4a^2b-a^3\right)c+ab^3-a^2b^2-a^ 3b+a^4}}\\ {{c^4+\left(-b-a\right)c^3+\left(-b^2+4ab-a^2 \right)c^2+\left(b^3+ab^2+a^2b+a^3\right)c}\over{ac^3+ \left(ab-a^2\right)c^2+\left(ab^2+4a^2b-a^3\right)c+ab ^3-a^2b^2-a^3b+a^4}} \end{bmatrix} $$ (OK, maxima doesn't really output like this, but you get the idea). So clearing denominator $$x=a\left(\sum_{cyc}a^2(a-b-c)+2bc(b+c)+4abc\right)$$ and cyclic permute for $y,z$:

factor(M . columnvector([

gives the output $[0,0,0]$.

Edit: correct factor of $2$, the expression is now much uglier.

  • $\begingroup$ No no (a, b, c) is not satisfying the equations, so there is another solution. And we need non trival solution. $\endgroup$ – nimmy Nov 14 '18 at 3:02
  • $\begingroup$ No no, I am sorry, once u please check a, b, c are not satisfying the equations $\endgroup$ – nimmy Nov 14 '18 at 5:37
  • $\begingroup$ No u typed the equation wrong $\endgroup$ – nimmy Nov 14 '18 at 5:40
  • $\begingroup$ U typed (b-c)*(2*s^3-2*s^2*a+s*(a^2-bc)+ab*c)..... $\endgroup$ – nimmy Nov 14 '18 at 5:41
  • $\begingroup$ Actually (b-c)*(2*s^3-2*s^2*a+s*(a^2-2bc)+ab*c)..... $\endgroup$ – nimmy Nov 14 '18 at 5:42

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