Question

9Which is the best name for a quadrilateral with vertices at A(5,-2), B(2,2), C(1,-5), and D(-2,-1)?A parallelogramB squarerhombusD rectangle

Answer 1

Parallelogram. Option A is correct

Explanations:

In order to determine the best name for a quadrilateral with the given vertices, we will find the measure of the distance AB, BC, CD, and AD using the distance formula as shown;

`D=\sqrt[]{(x_2-x_1)^2+(y_2-y^{}_1)^2}`

For the measure of AB with coordinates A(5,-2), B(2,2);

`\begin{gathered} AB=\sqrt[]{(5-2)^2+(-2-2^{}_{})^2} \\ AB=\sqrt[]{3^2+(-4)^2} \\ AB=\sqrt[]{9+16} \\ AB=\sqrt[]{25} \\ AB=5 \end{gathered}`

For the measure of BC with coordinates B(2,2) and C(1, -5)

`\begin{gathered} BC=\sqrt[]{(2-1)^2+(2-(-5^{}_{}))^2} \\ BC=\sqrt[]{1^2+7^2} \\ BC=\sqrt[]{50} \\ BC=5\sqrt[]{2} \end{gathered}`

For the measure of CD with coordinates C(1,-5), and D(-2,-1);

`\begin{gathered} CD=\sqrt[]{(1-(-2))^2+(-5-(-1^{}_{}))^2} \\ CD=\sqrt[]{3^2+(-4)^2} \\ CD=\sqrt[]{9+16} \\ CD=\sqrt[]{25} \\ CD=5 \end{gathered}`

For the measure of AD with coordinates A(5, -2), and D(-2,-1);

`\begin{gathered} AD=\sqrt[]{(5-(-2))^2+(-2-(-1^{}_{}))^2} \\ AD=\sqrt[]{(5+2)^2+(-2+1)^2} \\ AD=\sqrt[]{7^2+(-1)^2} \\ AD=\sqrt[]{50} \\ AD=5\sqrt[]{2} \end{gathered}`

For the slopes;

Check if the length AB is perpendicular to AD

`\begin{gathered} m_(AB)=(2+2)/(2-5) \\ m_(AB)=-(4)/(3) \end{gathered}`

For the slope of AD

`\begin{gathered} m_(AD)=(-1+2)/(-2-5) \\ m_(AD)=-(1)/(7) \end{gathered}`

Since AB is not perpendicular to AD, hence the quadrilateral is not a rectangle and also not a square or rhombus since all the sides are not equal.

From the given distances, you can see that opposite sides are equal (AB = CD and BC = AD ), hence the best name for a quadrilateral is a parallelogram.