Graph the parallclogram with vertices A (4,-6), B(-1,-6), C(3,-3), and D(8,-3).

Explanation
We are given the following coordinates:
[tex]\begin{gathered} A(4,-6) \\ B(-1,-6) \\ C(3,-3) \\ D(8,-3) \end{gathered}[/tex]We are required to graph the coordinates given and find the distance of the following:
[tex]\begin{gathered} AB \\ BC \\ CD \\ DA \end{gathered}[/tex]The points can be graphed as follows:
We know that the distance between two points can be calculated as:
Therefore, we can determine the distance of AB as:
[tex]\begin{gathered} A(4,-6)\to(x_1,y_1) \\ B(-1,-6)\to(x_2,y_2) \\ \therefore AB=\sqrt{(-1-4)^2+(-6-(-6))^2} \\ AB=\sqrt{(-1-4)^2+(-6+6)^2} \\ AB=\sqrt{(-5)^2+(0)^2} \\ AB=\sqrt{25+0}=\sqrt{25} \\ AB=5\text{ }units\text{ } \end{gathered}[/tex]We can determine the distance of BC as:
[tex]\begin{gathered} B(-1,-6)\to(x_1,y_1) \\ C(3,-3)\to(x_2,y_2) \\ \therefore BC=\sqrt{(3-(-1))^2+(-3-(-6))^2} \\ BC=\sqrt{(3+1)^2+(-3+6)^2} \\ BC=\sqrt{(4)^2+(3)^2} \\ BC=\sqrt{16+9}=\sqrt{25} \\ BC=5\text{ }units \end{gathered}[/tex]The value of CD is:
[tex]\begin{gathered} C(3,-3)\to(x_1,y_1) \\ D(8,-3)\to(x_2,y_2) \\ \therefore CD=\sqrt{(8-3)^2+(-3-(-3))^2} \\ CD=\sqrt{(8-3)^2+(-3+3)^2} \\ CD=\sqrt{(5)^2+(0)^2} \\ CD=\sqrt{25+0}=\sqrt{25} \\ CD=5\text{ }units \end{gathered}[/tex]Finally, the value of DA is:
[tex]\begin{gathered} D(8,-3)\to(x_1,y_1) \\ A(4,-6)\to(x_2,y_2) \\ \therefore DA=\sqrt{(4-8)^2+(-6-(-3))^2} \\ DA=\sqrt{(4-8)^2+(-6+3)^2} \\ DA=\sqrt{(-4)^2+(-3)^2} \\ DA=\sqrt{16+9}=\sqrt{25} \\ DA=5\text{ }units \end{gathered}[/tex]Hence, the answers are:
[tex]\begin{gathered} AB=5\text{ }units \\ BC=5\text{ }un\imaginaryI ts \\ CD=5\text{ }un\imaginaryI ts \\ DA=5\text{ }un\imaginaryI ts \end{gathered}[/tex]