Question

graph quadrilateral abcd using points a(-5, 6), b(-8, 2), c(-5, -2), and d(-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. b) determine the slope of each line segment in the quadrilateral. describe the relationship between the slopes. can you identify the figure? if so, identify the type of figure. if not, explain why not. ask yourself: what is the difference between a square and a rhombus?

Explanation:

Step1: Calculate side - length of AB

Use the distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For points $A(-5,6)$ and $B(-8,2)$, we have $x_1=-5,y_1 = 6,x_2=-8,y_2 = 2$. Then $AB=\sqrt{(-8 + 5)^2+(2 - 6)^2}=\sqrt{(-3)^2+(-4)^2}=\sqrt{9 + 16}=\sqrt{25}=5$.

Step2: Calculate side - length of BC

For points $B(-8,2)$ and $C(-5,-2)$, $x_1=-8,y_1 = 2,x_2=-5,y_2=-2$. Then $BC=\sqrt{(-5 + 8)^2+(-2 - 2)^2}=\sqrt{3^2+(-4)^2}=\sqrt{9 + 16}=\sqrt{25}=5$.

Step3: Calculate side - length of CD

For points $C(-5,-2)$ and $D(-2,2)$, $x_1=-5,y_1=-2,x_2=-2,y_2 = 2$. Then $CD=\sqrt{(-2 + 5)^2+(2 + 2)^2}=\sqrt{3^2+4^2}=\sqrt{9 + 16}=\sqrt{25}=5$.

Step4: Calculate side - length of DA

For points $D(-2,2)$ and $A(-5,6)$, $x_1=-2,y_1 = 2,x_2=-5,y_2 = 6$. Then $DA=\sqrt{(-5 + 2)^2+(6 - 2)^2}=\sqrt{(-3)^2+4^2}=\sqrt{9 + 16}=\sqrt{25}=5$.
Since all side - lengths are equal, it could be a rhombus or a square.

Step5: Calculate slope of AB

Use the slope formula $m=\frac{y_2 - y_1}{x_2 - x_1}$. For $A(-5,6)$ and $B(-8,2)$, $m_{AB}=\frac{2 - 6}{-8+5}=\frac{-4}{-3}=\frac{4}{3}$.

Step6: Calculate slope of BC

For $B(-8,2)$ and $C(-5,-2)$, $m_{BC}=\frac{-2 - 2}{-5 + 8}=\frac{-4}{3}$.

Step7: Calculate slope of CD

For $C(-5,-2)$ and $D(-2,2)$, $m_{CD}=\frac{2+2}{-2 + 5}=\frac{4}{3}$.

Step8: Calculate slope of DA

For $D(-2,2)$ and $A(-5,6)$, $m_{DA}=\frac{6 - 2}{-5+2}=\frac{4}{-3}=-\frac{4}{3}$.
Since the product of the slopes of adjacent sides $AB$ and $BC$ is $\frac{4}{3}\times(-\frac{4}{3})
eq - 1$, the figure is a rhombus.

Answer:

a. All side - lengths ($AB = BC=CD = DA = 5$). It is a rhombus.
b. $m_{AB}=\frac{4}{3}$, $m_{BC}=-\frac{4}{3}$, $m_{CD}=\frac{4}{3}$, $m_{DA}=-\frac{4}{3}$. Since the product of the slopes of adjacent sides is not - 1, the figure is a rhombus.