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Verify that parallelogram ABCD with vertices A(−5, −1), B(−9, 6), C(−1, 5), and D(3, −2)
is a rhombus by showing that it is a parallelogram with perpendicular diagonals.
(3 points)
To show that parallelogram ABCD is a rhombus, we first need to verify that it is a parallelogram. This can be done by showing that opposite sides are parallel.
Let's calculate the slope of each side of the parallelogram:
Slope of side AB:
m_AB = (y_B - y_A) / (x_B - x_A)
= (6 - (-1)) / (-9 - (-5))
= 7 / -4
= -7/4
Slope of side BC:
m_BC = (y_C - y_B) / (x_C - x_B)
= (5 - 6) / (-1 - (-9))
= -1 / 8
Slope of side CD:
m_CD = (y_D - y_C) / (x_D - x_C)
= (-2 - 5) / (3 - (-1))
= -7 / 4
Slope of side DA:
m_DA = (y_A - y_D) / (x_A - x_D)
= (-1 - (-2)) / (-5 - 3)
= 1 / 8
Since opposite sides AB and CD have the same slope (-7/4), and opposite sides BC and DA have the same slope (1/8), we can conclude that the parallelogram ABCD is indeed a parallelogram.
Next, we need to verify that the diagonals of the parallelogram are perpendicular. This can be done by showing that the product of the slopes of the diagonals is -1.
Slope of diagonal AC:
m_AC = (y_C - y_A) / (x_C - x_A)
= (5 - (-1)) / (-1 - (-5))
= 6 / 4
= 3/2
Slope of diagonal BD:
m_BD = (y_D - y_B) / (x_D - x_B)
= (-2 - 6) / (3 - (-9))
= -8 / 12
= -2/3
The product of the slopes of the diagonals AC and BD is (3/2) * (-2/3) = -1. Since the product is -1, we can conclude that the diagonals of the parallelogram ABCD are perpendicular.
Therefore, parallelogram ABCD with vertices A(-5, -1), B(-9, 6), C(-1, 5), and D(3, -2) is a rhombus.