Is there a geometric proof of this geometric interpretation of the Vandermonde determinant formula?
Solution 1:
If $h_k$ are the projections of $p_k$ onto $x$ axis, then we have: $$ A(p_1,p_2,p_3)=A(h_1,h_3,p_3,p_1)-A(h_1,h_2,p_2,p_1)-A(h_2,h_3,p_3,p_2), $$ where I assumed $x_1<x_2<x_3$ to avoid using absolute values. The areas of the three trapezoids on the r.h.s. can be computed and the result is: $$ A(p_1,p_2,p_3)= -{1\over2}\big[(x_1^2+x_3^2)(x_1-x_3)+(x_1^2+x_2^2)(x_2-x_1) +(x_2^2+x_3^2)(x_3-x_2)\big], $$ which is the same as the expression given in the question.
