the perimeter of a square is 4 times as great as the length of any of its sides. Determine whether the perimeter of the square is proportional to the side length. Explain your reasoning

Prove that the quadrilateral whose vertices are I(-2,3), J(2,6), K(7,6), and L(3, 3) is a rhombus.

I think in these problems the first step is to express each side as a vector.  A vector is the difference between points.  When two sides have the same vector (or negatives) it means they're parallel and congruent.  So in a rhombus IJKL the vectors IJ and LK should be the same, as should JK and IL.  That much assures a parallelogram; we check IJ and JK are congruent to complete the crowing of the rhombus.

Let's calculate these vectors:

IJ = J - I = (2,6) - (-2,3) = (2 - -2, 6 - 3) = (4, 3)

LK = K - L = (7, 6) - (3, 3) = (4, 3)

IJ = LK, so far so good

(Note: If you haven't got to vectors yet you can just show the two sides are the same length, 5, and have the same slope, 3/4, both of which can be read off the vectors.)

JK = K - J = (7,6) - (2,6) = (5,0)

IL = L - I = (3, 3) - (-2, 3)  = (5, 0)

Those are the same too.    

Now we have to show IJ ≅ JK

The length of IJ is the cliche √4²+3² = 5, the same as JK, so IJ ≅ JK

We showed all four sides are congruent and we have two pair of parallel sides, so we have a rhombus.