Precalculus, Chapter 1, 1.1, Section 1.1, Problem 34

Plot the points $A = (4,-3), B = (4,1)$ and $C = (2,1)$ and form the triangle $ABC$. Verify that the triangle is a right triangle. Determine its area.








$\begin{equation} \begin{aligned} AB =& \sqrt{(4-4)^2 + [1-(-3)]^2} \\ =& \sqrt{0+16} \\ =& \sqrt{16} \\ =& 4 \\ BC =& \sqrt{(2-4)^2 + (1-1)^2} \\ =& \sqrt{4+0} \\ =& \sqrt{4} \\ =& 2 \\ AC =& \sqrt{(2-4)^2 + [1-(-3)]^2} \\ =& \sqrt{4+16} \\ =& \sqrt{20} \\ =& 2 \sqrt{5} \\ (AC)^2 =& (AB)^2 + (BC)^2 \\ (2 \sqrt{5})^2 =& (4)^2 + (2)^2 \\ (4 \cdot 5) =& 16 + 4 \\ 20 =& 20 \end{aligned} \end{equation}$


Thus, $\Delta ABC$ is a right triangle.

The area of $\displaystyle \Delta ABC = \frac{1}{2} AB \cdot BC$,


$\begin{equation} \begin{aligned} \Delta ABC =& \frac{1}{2} (4)(2) \\ \\ =& \frac{8}{2} \\ \\ =& 4 \text{ square units} \end{aligned} \end{equation}$