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Question

Prove that (5,3),(1,11),(7,6),(1,2) coordinates are the vertices of parallelograms

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Solution

We know that the distance between the two points (x1,y1) and (x2,y2) is
d=(x2x1)2+(y2y1)2

Let the given vertices be A=(5,3), B=(1,11), C=(7,6) and D=(1,2)

We first find the distance between A=(5,3) and B=(1,11) as follows:

AB=(x2x1)2+(y2y1)2=(1(5))2+(11(3))2=(1+5)2+(11+3)2=62+(8)2
=36+64=100=10

Similarly, the distance between B=(1,11) and C=(7,6) is:

BC=(x2x1)2+(y2y1)2=(71)2+(6(11))2=62+(6+11)2=62+52=36+25
=61

Now, the distance between C=(7,6) and D=(1,2) is:

CD=(x2x1)2+(y2y1)2=(17)2+(2(6))2=(6)2+(2+6)2=(6)2+82=36+64
=100=10

Now, the distance between D=(1,2) and A=(5,3) is:

DA=(x2x1)2+(y2y1)2=(1(5))2+(2(3))2=(1+5)2+(2+3)2=62+52=36+25
=61

We also know that if the opposite sides have equal side lengths, then ABCD is a parallelogram.

Here, since the lengths of the opposite sides are equal that is:

AB=CD=10 and BC=DA=61

Hence, the given vertices are the vertices of parallelogram.

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