Question

Complete the proof to show that ABCD is a parallelogram.

On a coordinate plane, quadrilateral A B C D is shown. Point A is at (negative 2, negative 2), point B is at (negative 3, 4), point C is at (2, 2), and point D is at (3, negative 4).

The slope of Line segment B C is StartFraction 4 minus 2 Over negative 3 minus 2 EndFraction = negative two-fifths

The slope of Line segment A D is StartFraction negative 4 minus (negative 2) Over 3 minus (negative 2) EndFraction = StartFraction negative 4 + 2 Over 3 + 2 EndFraction = negative two-fifths

The slope of Line segment C D is StartFraction 2 minus (negative 4) Over 2 minus 3 EndFraction = StartFraction 2 + 4 Over 2 minus 3 EndFraction = StartFraction 6 Over negative 1 EndFraction = negative 6

The slope of Line segment B A is StartFraction 4 minus (negative 2) Over negative 3 minus (negative 2) EndFraction = StartFraction 4 + 2 Over negative 3 + 2 EndFraction = StartFraction 6 Over negative 1 EndFraction = negative 6

Answer 1

We have computed the slopes of the four sides:

- slope(BC) = (4 − 2)/(−3 − 2) = −2/5
- slope(AD) = (−4 − (−2))/(3 − (−2)) = (−2)/5 = −2/5

Thus slope(BC) = slope(AD), so BC ∥ AD.

- slope(CD) = (2 − (−4))/(2 − 3) = 6/(−1) = −6
- slope(BA) = (4 − (−2))/(−3 − (−2)) = 6/(−1) = −6

Thus slope(CD) = slope(BA), so CD ∥ BA.

Since both pairs of opposite sides are parallel, by definition ABCD is a parallelogram.