Question

Quadrilateral ABCD has vertices at A(3,−2), B(4,3), C(−2,0), and D(−3,−5). Quadrilateral A B C D in the plane with the given coordinates. © 2019 StrongMind. Created using GeoGebra. Which answers prove that quadrilateral ABCD is a parallelogram? Select all that apply. slope of AB¯¯¯¯¯¯¯¯=slope of DC¯¯¯¯¯¯¯¯; AD¯¯¯¯¯¯¯¯≅BC¯¯¯¯¯¯¯¯ AB¯¯¯¯¯¯¯¯∥DC¯¯¯¯¯¯¯¯ and AD¯¯¯¯¯¯¯¯∥BC¯¯¯¯¯¯¯¯ ∠A is supplementary to ∠B; ∠A≅∠C AD¯¯¯¯¯¯¯¯≅BC¯¯¯¯¯¯¯¯≅CD¯¯¯¯¯¯¯¯ slope of AD¯¯¯¯¯¯¯¯=slope of BC¯¯¯¯¯¯¯¯; AD¯¯¯¯¯¯¯¯≅BC¯¯¯¯¯¯¯¯

Answer 1

Answer:

The quadrilateral ABCD is a parallelogram as, slope of AB = slope of DC = 5, and slope of AD = slope of BC = 0.5.

Step-by-step explanation:

Quadrilateral ABCD has vertices at A(3,−2), B(4,3), C(−2,0), and D(−3,−5).

So, the slope of AB = $\frac{3 + 2}{4 - 3} = 5$

Now, the slope of BC = $\frac{0 - 3}{- 2 - 4} = \frac{1}{2}$

The slope of CD = $\frac{- 5 - 0}{- 3 + 2} = 5$

Finally, the slope of DA = $\frac{- 2 + 5}{3 + 3} = \frac{1}{2}$

Therefore, the quadrilateral ABCD is a parallelogram as, slope of AB = slope of DC = 5, and slope of AD = slope of BC = 0.5. (Answer)

Note: A quadrilateral to be a parallelogram the slope of each pair of opposite sides must be the same.