The vertices of quadrilateral ABCD are A(−5, 1) , B(−2, 5) , C(5, 3) , and D(2, −1) .
Slope of AB :
[tex]\mathrm{Slope\:between\:two\:points}:\quad \mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}[/tex]
[tex]m=\frac{5-1}{-2-\left(-5\right)}[/tex]
[tex]m=\frac{4}{3}[/tex]
Slope of BC :
[tex]m=\frac{3-5}{5-\left(-2\right)}=-\frac{2}{7}[/tex]
Slope of CD :
[tex]m=\frac{-1-3}{2-5}=\frac{4}{3}[/tex]
Slope of AD :
[tex]m=\frac{-1-1}{2-\left(-5\right)}=-\frac{2}{7}[/tex].
Therefore,
The slope of AB 1) 4/3, the slope of BC is 2) -2/7, the slope of CD is1) 4/3, and the slope of AD is 2) -2/7.
Quadrilateral ABCD is a 5) parallelogram because 8) both pairs of opposite sides are parallel.
Answer:
I took the test!! Have a great day!
Step-by-step explanation:
