The area of a triangle with vertices $A (3,0), B (7,0)$ and $C (8,4)$ is

If the vertices of $\Delta ABC$ are $A(-4, -1)$,$B(1, 2)$,and $C(4, -3)$,then the area of the triangle is:

$A$ median of a triangle divides it into two triangles of equal areas. Verify this result for $\Delta ABC$ whose vertices are $A (4, -6), B (3, -2)$ and $C (5, 2)$. Find the area of each triangle (in square units).

If $A(x_1, y_1), B(x_2, y_2),$ and $C(x_3, y_3)$ are the vertices of $\Delta ABC,$ then the area of $\Delta ABC$ is equal to:

Prove that the area of the triangle formed by joining the midpoints of the sides of a triangle is one-fourth the area of the original triangle.

Find the area of a triangle having the vertices $A(5, 3)$,$B(4, 5)$,and $C(3, 1)$.