Question

verify that parallelogram ABCD with vertices A(-5,-1) B (-9,6) C(-1,5) and D (3,-2) is a rhombus. do the following •identify the diagonals of the parallelogram •compute the slope of the diagonal •explain how you know the parallelogram is a rhombus

Answer 1

The diagonals of the parallelogram are A(-5, -1), C(-1, 5) and B(-9, 6), D(3, -2).
Slope of diagonal AC = (5 - (-1)) / (-1 - (-5)) = (5 + 1) / (-1 + 5) = 6 / 4 = 3/2
Slope of diagonal BD = (-2 - 6) / (3 - (-9)) = -8 / (3 + 9) = -8 / 12 = -2/3
For the parallelogram to be a rhombus, the intersection of the diagonals are perpendicular.
i.e. the product of the two slopes equals to -1.

Slope AC x slope BD = 3/2 x -2/3 = -1.
Therefore, the parallelogram is a rhombus.

Answer 2

Slopes of the diagonals are: 3/2 and -2/3 respectively.

Parallelogram ABCD is a rhombus, because the product of the slopes of its diagonals = -1.

What is the Slope of the Diagonals of a Rhombus?

Slope = change in y/change in x.

A rhombus has two diagonals, whereby, the product of the slopes = -1.

Given:

• A(-5,-1)
• B (-9,6)
• C(-1,5)
• D (3,-2)

Thus, lets find the slopes of diagonals AC and BD of parallelogram ABCD that is a rhombus:

Slope of AC = (5 - (-1))/(-1 -(-5)) = 6/4 = 3/2

Slope of BD = (-2 - 6)/(3 - (-9)) = -8/12 = -2/3

Product of the slopes of parallelogram ABCD = 3/2 × -2/3 = -1.

Therefore, parallelogram ABCD is a rhombus, because the product of the slopes of its diagonals = -1.

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