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

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.