Question

coordinate plane
in this activity, you will classify quadrilaterals by examining the lengths and relationships of their sides.
1 graph quadrilateral abcd using points a (-5, 6), b (-8, 2), c (-5, -2), and d (-2, 2).
2 consider the sides of quadrilateral abcd.
a determine each side length of quadrilateral abcd. can you classify quadrilateral abcd from its side lengths? if so, identify the type of figure. if not, explain why not.

Explanation:

Step1: Recall distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step2: Calculate length of side AB

For points $A(-5,6)$ and $B(-8,2)$, we have $x_1=-5,y_1 = 6,x_2=-8,y_2 = 2$.
\[

$$\begin{align*} AB&=\sqrt{(-8+ 5)^2+(2 - 6)^2}\\ &=\sqrt{(-3)^2+(-4)^2}\\ &=\sqrt{9 + 16}\\ &=\sqrt{25}\\ &=5 \end{align*}$$

\]

Step3: Calculate length of side BC

For points $B(-8,2)$ and $C(-5,-2)$, we have $x_1=-8,y_1 = 2,x_2=-5,y_2=-2$.
\[

$$\begin{align*} BC&=\sqrt{(-5 + 8)^2+(-2 - 2)^2}\\ &=\sqrt{3^2+(-4)^2}\\ &=\sqrt{9+16}\\ &=\sqrt{25}\\ &=5 \end{align*}$$

\]

Step4: Calculate length of side CD

For points $C(-5,-2)$ and $D(-2,2)$, we have $x_1=-5,y_1=-2,x_2=-2,y_2 = 2$.
\[

$$\begin{align*} CD&=\sqrt{(-2 + 5)^2+(2+2)^2}\\ &=\sqrt{3^2+4^2}\\ &=\sqrt{9 + 16}\\ &=\sqrt{25}\\ &=5 \end{align*}$$

\]

Step5: Calculate length of side DA

For points $D(-2,2)$ and $A(-5,6)$, we have $x_1=-2,y_1 = 2,x_2=-5,y_2 = 6$.
\[

$$\begin{align*} DA&=\sqrt{(-5 + 2)^2+(6 - 2)^2}\\ &=\sqrt{(-3)^2+4^2}\\ &=\sqrt{9+16}\\ &=\sqrt{25}\\ &=5 \end{align*}$$

\]
Since $AB = BC=CD=DA = 5$, the quadrilateral $ABCD$ is a rhombus.

Answer:

$AB = 5$, $BC = 5$, $CD = 5$, $DA = 5$. The quadrilateral $ABCD$ is a rhombus.