Question

WILL GIVE BRAINLIEST! Prove that a quadrilateral ABCD with vertices A(–1, –2), B( 2, –5), C(1, –2), and D( –2, 1) is a parallelogram. Find the point of intersection of the diagonals of parallelogram ABCD. Provide your complete solutions and proofs in your paper homework and enter the numeric answers online.

Answer 1

Answer:

Step-by-step explanation:

To prove a quadrilateral a parallelogram we prove,

1). Length of opposite sides are equal.

2). Slopes of the opposite sides are same.

Length of AB = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

                      = $\sqrt{(2+1)^2+(-5+2)^2}$

                      = $3\sqrt{2}$

Length of BC = $\sqrt{(2-1)^2+(-5+2)^2}$

                      = $\sqrt{10}$

Length of CD = $\sqrt{(1+2)^2+(-2-1)^2}$

                      = $3\sqrt{2}$

Length of AD = $\sqrt{(-1+2)^2+(-2-1)^2}$

                      = $\sqrt{10}$

Therefore, AB = CD and BC = AD (Opposite sides are equal in length)

Slope of AB = $\frac{y_2-y_1}{x_2-x_1}$

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

                    = -1

Slope of BC = $\frac{-5+2}{2-1}$

                    = -3

Slope of CD = $\frac{1+2}{-2-1}$

                    = -1

Slope of AD = $\frac{-2-1}{-1+2}$

                    = -3

Slope of AB = slope of CD and slope of BC = slope AD

Therefore, AB║CD and BC║AD

Hence ABCD is a parallelogram