Answer:
Trapezoid
Step-by-step explanation:
Given quadrilateral has vertices at points A(-2,-1), B(3,13), C(15,5) and D(13,-11).
Find slopes of lines AD and BC:
[tex]\text{Slope}_{AD}=\dfrac{y_D-y_A}{x_D-x_A}=\dfrac{-11-(-1)}{13-(-2)}=\dfrac{-11+1}{13+2}=\dfrac{-10}{15}=-\dfrac{2}{3}\\ \\\text{Slope}_{BC}=\dfrac{y_C-y_B}{x_C-x_B}=\dfrac{5-13}{15-3}=\dfrac{-8}{12}=-\dfrac{2}{3}[/tex]
Since the slopes are the same, lines AD and BC are parallel.
Find slopes of lines ABD and CD:
[tex]\text{Slope}_{AB}=\dfrac{y_B-y_A}{x_B-x_A}=\dfrac{13-(-1)}{3-(-2)}=\dfrac{14}{5}\\ \\\text{Slope}_{CD}=\dfrac{y_D-y_C}{x_D-x_C}=\dfrac{-11-5}{13-15}=\dfrac{-16}{-2}=8[/tex]
Since the slopes are different, lines AB and CD are not parallel.
This means quadrilateral ABCD is trapezoid (two opposite sides - parallel and two another opposite sides - not parallel)
