Q.If $(k, 2)$, $(2, 4)$ and $(3, 2)$ are vertices of the triangle of area $4$ square units then determine the value of $k$.

For a triangle with vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$, the familiar shoelace-type area formula

can be written compactly as a determinant:

The absolute value is taken because the raw determinant can come out negative depending on the (clockwise/anticlockwise) order in which the vertices are listed, whereas area is always non-negative.

Collinearity test. Three points are collinear (lie on one straight line) precisely when the triangle they would form has zero area, i.e.

This is one of the most common uses of a $3\times 3$ determinant in coordinate geometry: instead of computing slopes pairwise, a single determinant decides collinearity in one step, and (via row operations, Property 7) it often simplifies dramatically before you need to expand it at all.