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

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You also may use the information that diagonals of a parallelogram cuts each other in halves, hence, you may test if the midpoint of segment whose vertices are `(0,-1), (6,7)` is equal to the midpoint of segment whose vertices are `(-2,3), (8,3)` , such that:

`x_1 = (0+6)/2 => x = 3`

`y_1 = (-1+7)/2 => y = 3`

`x_2 = (-2+8)/2 => x_2 = 3`

`y_2 = (3+3)/2 => y_2 = 3`

Hence, evaluating the midpoints `(x_1,y_1)` and `(x_2,y_2)` yields that they represent the same point `(3,3)` , hence, the given quadrilater represents 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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