# What type of quadrilateral is the one that has the vertices A(-5,4), B(3,5), C(7,-2) ,D(-1,-3)?

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For the given quadrilateral ABCD:

Length of AB = sqrt(8^2 + 1^2) = sqrt 65, slope of AB = (1/8)

Length of BC = sqrt(4^2 + 7^2) = sqrt 65, slope of BC = -7/4

Length of CD = sqrt(8^2 + 1^2) = sqrt 65, slope of CD = 1/8

Length of DA = sqrt(7^2 + 4^2) = sqrt 65, slope of DA = -7/4

So we have a quadrilateral with all sides equal and opposite sides parallel to each other. But the adjacent sides are not perpendicular to each other.

**The quadrilateral ABCD is a rhombus.**

If the opposite sides of the given quadrilateral are parallel, then the quadrilateral is parallelogram.

We'll compute the slopes of the opposite sides AB and CD:

mAB = (yB - yA)/(xB - xA)

mAB = (5-4)/(3+5)

mAB =1/8

mCD = (yD - yC)/(xD - xC)

mCD = (-3+2)/(-1-7)

mCD = -1/-8

mCD = 1/8

Since the values of the slopes are equal, then the opposite sides AB and CD are parallel.

We'll verify if the slopes of AD and BC are parallel, too.

mBC = (-2-5)/(7-3)

mBC = -7/4

mAD = (-3-4)/(-1+5)

mAD = -7/4

Since the values of the slopes are equal, then the opposite sides AD and BC are parallel.

Now, we'll verify if the parallelogram is rectangle.

We'll calculate the product of the supposed perpendicular lines:

mAB*mBC = -7/32

We notice that the value of the product is not -1, then the parallelogram is not a rectangle.

We'll verify if the parallelogram is a rhombus.

We'll calculate the lengths of two consecutive sides.

[AB] = sqrt[(3+5)^2 + (5-4)^2]

[AB] = sqrt(64 + 1)

[AB] = sqrt 65

[BC] = sqrt[(7-3)^2 + (-2-5)^2]

[BC] = sqrt(16+49)

[BC] = sqrt 65

**Since the lengths of two consecutive sides are equal, then the quadrilateral ABCD is a rhombus.**