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Question

If two vertices of a parellelogram are (3,2) and (1,0) and the diagonals intersect at (2,5), then the other two vertices are:

A
(1,10),(5,12)
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B
(1,12),(5,10)
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C
(2,10),(5,12)
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D
(1,10),(2,12)
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Solution

The correct option is D (1,12),(5,10)
Let the given vertices be A(1,12)B(5,10)C(a,b);D(x,y)
We know that the diagonals of a parallelogram bisect each other. So,the midpoint of AC =(2,5) and of BD
$ = (2,-5) $.
Mid point of two points (x1,y1) and (x2,y2) is calculated by the formula
(x1+x22,y1+y22)
Hence,
midpoint of AC=(2,5)
=>(1+a2,12+b2)=(2,5)
=>1+a2=2;12+b2=5
=>1+a=4;12+b=10
a=3;b=2
Hence, C =(3,2)
And, midpoint of BD=(2,5)
=>(5+x2,10+y2)=(2,5)
=>5+x2=2;10+y2=5
=>5+x=4;10+y=10
x=1;y=0
=>D=(1,0)

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