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 = [tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
= [tex]\sqrt{(2+1)^2+(-5+2)^2}[/tex]
= [tex]3\sqrt{2}[/tex]
Length of BC = [tex]\sqrt{(2-1)^2+(-5+2)^2}[/tex]
= [tex]\sqrt{10}[/tex]
Length of CD = [tex]\sqrt{(1+2)^2+(-2-1)^2}[/tex]
= [tex]3\sqrt{2}[/tex]
Length of AD = [tex]\sqrt{(-1+2)^2+(-2-1)^2}[/tex]
= [tex]\sqrt{10}[/tex]
Therefore, AB = CD and BC = AD (Opposite sides are equal in length)
Slope of AB = [tex]\frac{y_2-y_1}{x_2-x_1}[/tex]
= [tex]\frac{-5+2}{2+1}[/tex]
= -1
Slope of BC = [tex]\frac{-5+2}{2-1}[/tex]
= -3
Slope of CD = [tex]\frac{1+2}{-2-1}[/tex]
= -1
Slope of AD = [tex]\frac{-2-1}{-1+2}[/tex]
= -3
Slope of AB = slope of CD and slope of BC = slope AD
Therefore, AB║CD and BC║AD
Hence ABCD is a parallelogram
