Q.Find the area of the triangle with vertices $(1,0),(6,0),(4,3)$ using determinants.

Given a triangle's vertices $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$, its area is

derived by splitting the triangle's area into three trapeziums against the x-axis and simplifying. Because listing the vertices in a different rotational order swaps two rows of the determinant (Property 2), the signed determinant can come out negative — so the area itself is always taken as the absolute value. This immediately gives a clean collinearity test: if the same determinant evaluates to exactly $0$, the "triangle" has zero area, which can only happen when the three points lie on one straight line. This formula and its collinearity corollary are used constantly in coordinate geometry, and the same determinant reappears (with the constant column swapped in) as the very consistency-condition determinant of §4.3.2, since three collinear points are really a degenerate case of three "equations" (through pairs of the points) all agreeing.