Answer:
Step-by-step explanation:
Given that coordinates of the vertices of quadrilateral ABCD are A(-1, -1), B(-3, 3), C(1, 5), and D(5, 2).
Slope of line is given as
[tex] slope = \frac{ Yj - Yi }{ Xj - Xi } [/tex]
Using above formula,
The slope of AB is [tex] S1 =\frac{ Yj - Yi }{ Xj - Xi } [/tex]
[tex] S1 = \frac{3-(-1)}{(-3)-(-1)} [/tex]
[tex] S1 = \frac{4}{-2} = -2 [/tex]
The slope of BC is [tex] S2 = \frac{ Yj - Yi }{ Xj - Xi } [/tex]
[tex] S2 = \frac{5-3}{1-(-3)} [/tex]
[tex] S2 = \frac{2}{4} = \frac{1}{2} [/tex]
The slope of CD is [tex] S3 = \frac{ Yj - Yi }{ Xj - Xi } [/tex]
[tex] S3 = \frac{2-5}{5-1} [/tex]
[tex] S3 = \frac{-3}{4} [/tex]
The slope of AD is [tex] S4 = \frac{ Yj - Yi }{ Xj - Xi } [/tex]
[tex] S3 = \frac{(-1) - 2}{(-1) - 5} [/tex]
[tex] S3 = \frac{-3}{-6} = \frac{1}{2} [/tex]
You can see that Only Line BC and Line AD are parallel.
Therefore, Quadrilateral ABCD is a Trapezoid.
