# What is the area of the polygon with the following vertices? (4,1),(3,4),(−3,2),(−2,−1) simple please & thank you

Jun 23, 2018

Area: $\textcolor{b l u e}{20 \text{ sq.units}}$

#### Explanation:

For ease of reference I will label the vertices:
$\textcolor{w h i t e}{\text{XXX}} A : \left(4 , 1\right)$
$\textcolor{w h i t e}{\text{XXX}} B : \left(3 , 4\right)$
$\textcolor{w h i t e}{\text{XXX}} C : \left(- 3 , 2\right)$
$\textcolor{w h i t e}{\text{XXX}} D : \left(- 2 , - 1\right)$

Notice that the slopes of $A B$ and $C D$ are both $- 3$
$\textcolor{w h i t e}{\text{XXX}}$You can determine this, for example, for $A B$ by
$\textcolor{w h i t e}{\text{XXX}} {m}_{A B} = \frac{\Delta {y}_{A B}}{\Delta {x}_{A B}} = \frac{4 - 1}{3 - 4} = - 3$

Similarly we can note that the slopes of $B C$ and $D A$ are both $\frac{1}{3}$.

Since the slopes of $A B$ and $C D$ are negative reciprocals of $B C$ and $D A$,
$A B$ and $C D$ are perpendicular to $B C$ and $D A$.
$\Rightarrow A B C D$ is a rectangle.

The area of $A B C D$ can be calculated as the length $\left\mid A B \right\mid$ ties the length $\left\mid C D \right\mid$

Using the Pythagorean Theorem:
$\left\mid A B \right\mid = \sqrt{\Delta {x}_{A B}^{2} + \Delta {y}_{A B}^{2}}$
$\textcolor{w h i t e}{\left\mid A B \right\mid} = \sqrt{{1}^{1} + {3}^{2}} = \sqrt{10}$
and
$\left\mid C D \right\mid = \sqrt{{\Delta}_{C D} {x}^{2} + \Delta {y}_{C D}^{2}}$
$\textcolor{w h i t e}{\left\mid C D \right\mid} = \sqrt{{2}^{2} _ {6}^{2}} = \sqrt{40} = 2 \sqrt{10}$

Therefore
${\text{Area}}_{A B C D} = \sqrt{10} \times 2 \sqrt{10} = 20$