Quadrilateral abcd has coordinates a (3, 1), b (4, 4), c (7, 5), d (6, 2). quadrilateral abcd is a?

Here, we need to calculate the distance between the point by applying formula:
d = √(y2-y1)² + (x2-x1)²

After applying the formula, we get, 
AB=BC=CD=DA and BD !=AC

So, It is a property of Rhombus. Hence, It is a Rhombus

Hope this helps!

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step-by-step explanation:

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