Question

What is the area of $\text{∆ABC}$ with vertices $A(2,3),B(1,-3),$and $C(-3,1)$?

Answer 1

Find the area of the $\text{∆ABC}$:

The vertices are $A(2,3),B(1,-3),$and $C(-3,1)$.

The area of the triangle with vertices is given by, $A=\left|\left(\frac{1}{2}\right)\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\right|$.
Here, $\left(x_1,y_1\right)=(2,3),\left(x_2,y_2\right)=(1,-3),\left(x_3,y_3\right)=(-3,1)$

Now, Substitute the value of the given point with the above formula:

$\begin{array}{rcl}A& =& \left|\frac{1}{2}\left[2\left(-3-1\right)+1\left(1-3\right)+\left(-3\right)\left\{3-\left(-3\right)\right\}\right]\right|\\ & =& \left|\frac{1}{2}\left[2\left(-4\right)+1\left(-2\right)+\left\{\left(-3\right)\left(6\right)\right\}\right]\right|\\ & =& \left|\frac{1}{2}\left[-8-2-18\right]\right|\\ & =& \left|\frac{1}{2}\left[-28\right]\right|\\ & =& 14\end{array}$

Therefore, the area of the $\text{∆ABC}$ is $14$ square unit.