Question

Prove that the points(3, 2), (4,0), (6, -3) and
(5, -5) are the vertices of a parallelogram.​

Answer 1

Answer:

Step-by-step explanation:

When you copy the task, please be precise, because (3, 2) is not a vertex of given parallelogram.

Let A(3, -2), B(4, 0), C(6, - 3), D(5, - 5). Slopes of parallel lines are equal.

$m_{AB}$ = (0 - (-2)) / (4 - 3) = 2

$m_{CD}$ = (- 5 - (- 3)) / (5 - 6)  = 2

AB = √[(3-4)²+(-2-0)²] = √5

CD = √[(6-5)²+(-3-(-5))²] = √5

$m_{BC}$ = (- 3 - 0) / (6 - 4) = - 3/2

$m_{AD}$ = (- 5 - (- 2)) / (5 - 3) = - 3/2

BC = √[(-3-0)²+(6-4)²] = √13

AD = √[(-5-(-2))²+(5-3)²] = √13

AB║CD and AB = CD, BC║AD and BC = AD ⇒ ABCD is parallelogram.