Parallelogram abcd has vertices at a(0, 0) , b(3, 6) , c(5, 5) , and d(2,−1) . which conclusion can be made?

A = ( 0 , 0 ), B = ( 3 , 6 ), C = ( 5 , 5 ) and B = ( 2 , −1 )

AB = √[(3-0)^2 + (6 - 0)^2] = √(9 + 36) = √45

BC = √[(3-5)^2 + (6 - 5)^2] = √(4 + 1) = √5

CD = √[(2-5)^2 + (-1 - 5)^2] = √(9 + 36) = √45

AD = √[(2-0)^2 + (-1 - 0)^2] = √(4 + 1) = √5

AB = CD = √45 and BC = AD = √5 ...So this can't be a square or rhombus because square and rhombus have 4 equal sides. It's a rectangle.

Slope of AB = 6/3 = 2

Slope of BC = (6 - 5) / (3 -5) = -1/2

Both slopes are opposite and reciprocal so both lines are perpendicular(AB ⊥ BC)

Answer is the last option

AB ⊥ BC; therefore, ABCD is a rectangle.

Hope it helps.

AC=BD; therefore, ABCD is a rectangle

Step-by-step explanation:

2 x^2 -3x+p

sum of roots = 3/2

let other root be x

3+ x = 3/2

x= 3/2 - 3 = -3/2

product of roots = p/2

3× -3/2 = p/2

-9 = p

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the answer to your question a or b but i am pretty sure it is b. i hope this : )

step-by-step explanation: