Question

Quadrilateral WXYZ has vertices W(-4,-3), X(0,-1). Y(6,-2), Z(2,-4). Using properties of diagonals prove that WXYZ is a parallelogram, but NOT a rectangle.

Answer 1

The diagonals are bisectors of each other and meet at (1, -5/2) but are not congruent, so WXYZ is a parallelogram and not a rectangle

Explanation:

for parallelograms.

the diagonals intersect at the midpoints

W X

Z Y

Diagonals are WY and XZ

WY : (-4, -3) to (6, -2)

midpoint is [(-4 + 6)/2 , (-3 - 2)/2 ] = ( 1, -5/2 )

XZ : (0, -1) to (2, -4)

midpoint is [ (0 + 2)/2 , (-1 + -4)/2 ] = (1, -5/2)

The diagonals are bisectors of eachother.

Next show that these diagonals are not congruent.

length of WY = root ( (-4 - 6)^2 + (-3 - (-2))^2 )

WY = root ( 100 + 1)

WY = root(101)

length of XZ = root ( (0 - 2)^2 + ( -1 - -4)^2 )

XZ = root (4 + 9) = root (13)

we see that WY is not equal to XZ

...

so .. quadrilateral WXYZ is a parallelogram but not a rectangle.