Respuesta :
Answer:
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;
[tex]D=\sqrt[]{(x_2-x_1)^2+(y_2-y^{}_1)^2}[/tex]For the measure of AB with coordinates A(5,-2), B(2,2);
[tex]\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}[/tex]For the measure of BC with coordinates B(2,2) and C(1, -5)
[tex]\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}[/tex]For the measure of CD with coordinates C(1,-5), and D(-2,-1);
[tex]\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}[/tex]For the measure of AD with coordinates A(5, -2), and D(-2,-1);
[tex]\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}[/tex]For the slopes;
Check if the length AB is perpendicular to AD
[tex]\begin{gathered} m_{AB}=\frac{2+2}{2-5} \\ m_{AB}=-\frac{4}{3} \end{gathered}[/tex]For the slope of AD
[tex]\begin{gathered} m_{AD}=\frac{-1+2}{-2-5} \\ m_{AD}=-\frac{1}{7} \end{gathered}[/tex]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.
