Given the quadrilateral PQRS
As shown the quadrilateral has the following vertices
P(-1, -7), Q(6, -4), R(2, 5), and S(-5, 2)
We will prove the quadrilateral is a parallelogram by finding the length of the sides using the following formula:
[tex]$d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}$[/tex]
(a) we will find the length of PS and QR
[tex]PS=\sqrt{(-5-(-1))^2+(2-(-7)^2}=\sqrt{(-4)^2+(9)^2}=\sqrt{97}[/tex][tex]QR=\sqrt{(2-6)^2+(5-(-4))^2}=\sqrt{(-4)^2+(9)^2}=\sqrt{97}[/tex]
(b) we will find the slope of PS and QR
[tex]slope\text{ }of\text{ }PS=\frac{2--7}{-5--1}=\frac{9}{-4}[/tex][tex]slope\text{ }of\text{ }QR=\frac{5--4}{2-6}=\frac{9}{-4}[/tex]
(c) From parts (a) and (b), we can conclude option 2
the quadrilateral is a parallelogram because it has one pair of opposite sides that are both congruent and parallel
