Question

Find the area of the triangle whose vertices are $(8,4),(6,6)$ and $(3,9)$.

Answer 1

Let the $\Delta ABC$ have vertices $A(x_1,y_1)=(8,4),B(x_2,y_2)=(6,6)$ and $C(x_3,y_3)=(3,9)$.

Now $ar\Delta ABC$ with vertices $A(x_1,y_1),B(x_2,y_2)$ and $C(x_3,y_3)$ is

$ar\Delta ABC=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$\therefore ar\Delta ABC=\frac{1}{2}[8(6-9)+6(9-4)+3(4-6)]$ units

$\Rightarrow ar\Delta ABC=0$

Therefore, $ar\Delta ABC=0$

The points $A,B$ and $C$ are collinear, since $ar\Delta ABC=0$.