Show that the points (0;-1), (-2;3), (6;7), (8;3) are the vertices of a parallelogram.

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neela | High School Teacher | (Level 3) Valedictorian

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Let A(0,-1) , B(-2,3) , C(6,7) and D (8,3).

AB^2 = (-2-0)^2+(3- -1)^2 = 4+4^2 = 20.

BC^2 = (6- -2)^2 +(7-3)^2 = 8^2+4^2 = 80.

CD^2 = (8-6)^2 +(3-7)^2 =  4+4^2 = 20.

AD^2 = (0-8)^2 +(-3-1)^2 = 64+16 = 80.

Therefore AB^2 = 20 = CD^2

BC^2 = 80 =  A^2.

That proves that the the opposite pairs of sides are equal in length. . Therefore ABCD is a parallelogram.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To prove the the given points represent the vertices of a parallelogram, we'll have to compute the length of the sides that pass through the given vertices and if 2 of them are equal and the other 2 are also equal, then the geomwtric shape is a parallelogram.

We'll calculate the length of the side AB, that passes through the points (0;-1), (-2;3).

AB = sqrt [(-2-0)^2 + (3+1)^2]

AB = sqrt (4+16)

AB = 2sqrt5

We'll calculate the length of the side CD, that passes through the points (6;7), (8;3).

CD = sqrt[(8-6)^2 + (3-7)^2]

CD = sqrt (4 + 16)

CD = 2sqrt5

We remark that AB = CD = 2sqrt5.

We'll compute the length of BC:

BC = sqrt[(6+2)^2 + (7-3)^2]

BC = sqrt(64+16)

BC = 4sqrt5

We'll compute the length of AD:

AD = sqrt[(8-0)^2 + (3+1)^2]

AD = sqrt(64+16)

AD = 4sqrt5

We notice that the lengths of the sides AD and BC are equal to 4sqrt5.

The geometric shape whose vertices are (0;-1), (-2;3), (6;7), (8;3) is parallelogram.

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