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# Show that A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4) Are the Vertices of a Rhombus Abcd. - Geometry

Show that A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4) are the vertices of a
rhombus ABCD.

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#### Solution

A(-4, -7), B(-1, 2), C(8, 5), D(5, -4)
According to distance formula,
AB=sqrt([-1-(-4)]^2 +[2-(-7)]^2
∴ AB = sqrt(3^2+9^2)
∴ AB =sqrt(9+81)
∴ AB = sqrt90                      ...... (1)

BC =sqrt([8-(-1)]^2+(5-2)^2)
∴ BC =sqrt(9^2+3^2)
∴ BC = sqrt(81+9)
∴ BC = sqrt90                        ...... (2)

CD =sqrt((5-8)^2 +(-4-5)^2)
∴ CD =sqrt((-3)^2 +(-9)^2)

∴ CD =sqrt(9+81)

∴ CD = sqrt90           ....... (3)

AD = sqrt([5-(-4)]^2+ [-4-(-7)]^2)
∴ AD =sqrt(9^2+3^2)
∴ AD =sqrt(81+9)
∴ AD = sqrt90                      ....... (4)
From (1), (2), (3) and (4)
AB = BC = CD = AD
∴ square ABCD is a rhombus.

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