Answer:
Rhombus
Explanation:
The diagonals of a parallelogram bisect each other. That is, they have the same midpoint. If they are the same length, that parallelogram is a rectangle. If they cross at right angles, it is a rhombus. If it is a rectangle and rhombus, then it is a square.
Diagonal midpoints
The midpoints of AC and BD are ...
(A+C)/2 and (B+D)/2
To determine if the midpoints are the same, we can skip the division by 2 and simply look at the sums:
A +C = (3, 5) +(6, 2) = (9, 7)
B +D = (7, 6) +(2, 1) = (9, 7)
The midpoints of the diagonals are the same, so the figure is at least a parallelogram.
Diagonal vectors
The diagonal vectors will be the same length if the figure is a rectangle. They will be perpendicular if the figure is a rhombus. The vectors are ...
AC = C -A = (6, 2) -(3, 5) = (3, -3)
BD = D -B = (2, 1) -(7, 6) = (-5, -5)
The length of each of these is the root of the sum of squares of its components. These are obviously different lengths (3√2 vs 5√2).
The dot-product of these will be zero if they are perpendicular:
AC·BD = x1·x2 +y1·y2 = (3)(-5) +(-3)(-5) = -15 +15 = 0
Conclusion
The diagonals are different length and mutual perpendicular bisectors, so the figure is a rhombus.
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Additional comment
Looking at the dot-product is a simple way to check that the slopes are opposite reciprocals. The slope of a vector with components (x, y) is m = y/x.
The requirement that slopes be opposite reciprocals means ...
y1/x1 = -1/(y2/x2) . . . . . . . . slope relationship
(y1)(y2)/((x1)(x2)) = -1 . . . . . multiply by y2/x2
y1·y2 = -x1·x2 . . . . . . . . . . multiply by (x1·x2)
x1·x2 +y1·y2 = 0 . . . . . . . . add x1·x2
This shows the vector dot product being zero is equivalent to the slopes being opposite reciprocals. The vectors are perpendicular in this case.