If a(2, −1), b(5, −5), c(10, −5), and d(7, −1) are the vertices of a quadrilateral, do the points form a rhombus? justify your answer. a)no, it is not a rhombus because the slope of diagonal segment bd equals two and the slope of diagonal segment ac equals negative start fraction one over two end fraction full stop. b) yes, it is a rhombus because the slope of diagonal segment bd equals two, and the slope of diagonal segment ac equals negative two full stop, and the diagonals bisect each other. c) yes, it is a rhombus because the midpoint of diagonals segment bd and segment ac is (6, −3). d)yes, it is a rhombus because the slope of diagonal segment bd equals two, the slope of diagonal segment ac equals negative start fraction one over two end fraction, and the diagonals bisect each other.

Option A: establishes that the diagonals intersect at right angles, which is a property of a rhombus, but not sufficient.  Conclusion of "no" is incorrect.
Option B: establishes that the diagonals intersect at right angles, which is a property of a rhombus, but not sufficient.
Option C: establishes that the diagonals bisect each other, which is a property of a rhombus, but not sufficient.
Option D: establishes that the diagonals intersect at right angles, which is a property of a rhombus, AND diagonals bisect each other.  Together, the two properties are sufficient to establish that the figure is a rhombus.

Another sufficient condition to establish a rhombus is that the four sides are congruent, as follows:

If A(2, −1), B(5, −5), C(10, −5), and D(7, −1)  are the vertices of a rhombus IN ORDER, then AB, BC, CD, DA are the sides.

It is sufficient to prove that the four sides are congruent to conclude that ABCD is a rhombus.  (can be established by joining diagonals and proving congruence of triangles by SSS).

mAB=sqrt((5-2)^2+(-5-(-1))^2)=sqrt(3^2+4^2)=5
similarly,
mBC=sqrt(5^2+0)=5
mCD=sqrt(3^2+4^2)=5
mDA=sqrt(5^2+0)=5
Since all four sides are congruent, the figure is a rhombus.

5\16 x 2\5 = 10/80simplify to 1/80