I'm not sure how to solve this problem. Can you guys provide some input/hints?
Let $A=(x_1,y_1)$, $B=(x_2,y_2)$ and $C=(x_3,y_3)$ be three points in $\mathbb{R}^{2}$.
Show that the area of $\Delta ABC$ is given by
$\frac{1}{2}\left| \det(M) \right|$, where
$$M = \begin{pmatrix} 1 & 1
& 1\\ x_1 & x_2 & x_3\\ y_1 & y_2 & y_3 \end{pmatrix}$$
Best Answer
Another possibility is to use the formal properties of the determinant and see how they correspond to the properties of the area. This seems lengthy, but it explains also why you have such a relation between determinant and area.
You start with the determinant: $$ \left|\begin{array}{ccc} 1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{array} \right| $$ then you can subtract from the second row $x_1$ times the first row and from the third row you can subtract $y_1$ times the first row to get: $$ \left|\begin{array}{ccc} 1 & 1 & 1 \\ 0 & x_2-x_1 & x_3-x_1 \\ 0 & y_2-y_1 & y_3-y_1 \end{array} \right|. $$ This operation does not change the determinant (which is multilinear). And it does not change the area of the triangle, because it corresponds to a translation, namely the translation of vector $-(x_1,y_1)$ which sends the first vertex of the triangle to the origin.
Now suppose to fix ideas that $0<x_3<x_2$ and consider the triangle (which now has vertices: $p_1= (0,0)$, $p_2=(x_2-x_1,y_2-y_1)$, $p_3=(x_3-x_1, y_3-y_1)$) divided in two by the line $x=x_3-x_1$, and let $p$ be the point with $x$ coordinate equal to $p_3$ which lies on the edge $p_1 p_2$. If we consider $pp_3$ as the base of the two smaller triangles, you see that you can move the vertices $p_2$ and $p_1$ vertically (which means parallel to the common base $pp_3$ without changing the area of the resulting quadrilateral (which is the union of the two smaller triangles). Or, which is the same, you can keep $p_1$ fixed and move the base $pp_3$ vertially. If you want to move the triangle so that both the point $p$ and $p_2$ go to the $x$-axis, you find the mapping $$ \begin{cases} x' = x\\ y' = y - \frac{y_2-y_1}{x_2-x_1}x \end{cases} $$ which correspond to subtract from the third line of the matrix ($y$ coordinates) a multiple of the second line ($x$ coordinate). So, again, this transformation maintains the area of the triangle, and the determinant of the matrix.
You get: $$ \left|\begin{array}{ccc} 1 & 1 & 1 \\ 0 & x_2-x_1 & x_3-x_1 \\ 0 & 0 & (y_3-y_1)-\frac{y_2-y_1}{x_2-x_1}(x_3-x_1) \end{array} \right|. $$ So now you have a triangle (with same area of the original one) whose basis is on the $x$ axis has length $x_2-x_1$ and whose height is given by the $y$ coordinate of the third point which is the entry in the lower-right corner of the matrix. So the area is half the product of the entries on the diagonal, and in fact in this case (triangular matrix) this is indeed the determinant of the matrix.
You could also complete the diagonalization process, by moving the third vertex to the $y$ axis. In that case you would get a right triangle and a diagonal matrix.