M(-5,1); N(-6,-3); 0(-2,-2); P(-1,2)
Prove that MNOP is a Rhombus.

please help.

Explanation:

The quadrilateral will be a rhombus if the diagonals bisect each other and cross at right angles.

The diagonals will have the same midpoint if the sum of end-point coordinates is the same for each.

  M+O = N+P

  (-5, 1) +(-2, -2) = (-6, -3) +(-1, 2)

  (-5-2, 1-2) = (-6-1, -3+2)

  (-7, -1) = (-7, -1) . . . . . true, so diagonals bisect each other

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The diagonals will be at right angles if the product of their slopes is -1. The slope of each is ∆y/∆x.

  slope of MO = ∆y/∆x = (-2-1)/(-2-(-5)) = -3/3 = -1

  slope of NP = ∆y/∆x = (2-(-3), -1-(-6)) = 5/5 = 1

The product of the slopes is (-1)(1) = -1, so the diagonals are at right angles.

The diagonals bisect each other at right angles, so MNOP is a rhombus.

the square root of 16 is 4.

4*4 = 16