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Question

Prove that (4,1),(6,0),(7,2) and (5,1) are the vertices of a rhombus. Is it a square?

Solution
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Let the given points be A, B, C and D respectively. Then,Coordinates of the mid-point of Ac are(4+72,1+22)=(112,12)Coordinates of the mid-point of BD are (6+52,0+12)=(112,12)Thus, AC and BD have the same mid-point.
Hence, ABCD is a parallelogram.
Now,AB=(64)2+(0+1)2=5,BC=(76)2+(20)2=5AB=BCSo, ABCD is a parallelogram whose adjacent sides are equal.Hence,ABCD is a rhombus.
We have,AC=(74)2+(2+1)2=32,and,BD=(65)2+(01)2=2
Clearly, ACBD.So, ABCD is not a square.

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