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Question

If the mid points of sides of ΔABC are (2,3),(3,7) and (2,1), then the possible coordinates of its vertices are

A
(1,5)
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B
(3,3)
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C
(7,9)
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D
(5,1)
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Solution

The correct option is C (7,9)
Let A(x1,y1),B(x2,y2),C(x3,y3)
M1(2,3) is midpoint of side BC
M2(3,7) is midpoint of side AC
M3(2,1) is midpoint of side AB

Now, x1+x22=2
x1+x2=4 (1)
Similarly, x2+x3=4 (2)
x1+x3=6 (3)

Adding equation (1),(2) and (3),
2(x1+x2+x3)=6x1+x2+x3=3 (4)
Using equation (1) and (4),
x3=7
Using equation (2) and (4),
x1=1
Using equation (3) and (4),
x2=3

Similarly, for ycoordinates
y1+y2=2 (5)y2+y3=6 (6)y3+y1=14 (7)

Adding equation (5),(6) and (7)
y1+y2+y3=11 (8)
Using equation (5) and (8),
y3=9
Using equation (6) and (8),
y1=5
Using equation (7) and (8),
y2=3

Hence, the coordinates of the triangle vertex are
A=(1,5), B=(3,3), C=(7,9)


Alternate Solution :


Let A(x1,y1),B(x2,y2),C(x3,y3)
M1(2,3) is midpoint of side BC
M2(3,7) is midpoint of side AC
M3(2,1) is midpoint of side AB

We know that, BM1M2M3 is a parallelogram.
And its diagonal bisect each other,
(x2+32,y2+72)=(222,3+12)x2=3,y2=3B=(3,3)

Similarly, CM1M3M2 is a parallelogram.
(x322,y3+12)=(2+32,3+72)x3=7,y3=9C=(7,9)

Similarly, AM3M1M2 is a parallelogram.
(x1+22,y1+32)=(322,7+12)x1=1,y1=5A=(1,5)

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