Question

The coordinates of the vertices of quadrilateral ABCD are A(-1, -1), B(-3, 3), C(1, 5), and D(5, 2).

Answer 1

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

$slope = \frac{ Yj - Yi }{ Xj - Xi }$

Using above formula,

The slope of AB is $S1 =\frac{ Yj - Yi }{ Xj - Xi }$

$S1 = \frac{3-(-1)}{(-3)-(-1)}$

$S1 = \frac{4}{-2} = -2$

The slope of BC is $S2 = \frac{ Yj - Yi }{ Xj - Xi }$

$S2 = \frac{5-3}{1-(-3)}$

$S2 = \frac{2}{4} = \frac{1}{2}$

The slope of CD is $S3 = \frac{ Yj - Yi }{ Xj - Xi }$

$S3 = \frac{2-5}{5-1}$

$S3 = \frac{-3}{4}$

The slope of AD is $S4 = \frac{ Yj - Yi }{ Xj - Xi }$

$S3 = \frac{(-1) - 2}{(-1) - 5}$

$S3 = \frac{-3}{-6} = \frac{1}{2}$

You can see that Only Line BC and Line AD are parallel.

Therefore, Quadrilateral ABCD is a Trapezoid.