is a point such that where is point with coordinates .The least value of for which such point P exists is
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k = ( x − x 1 ) 2 + ( y − y 1 ) 2 + ( x − x 2 ) 2 + ( y − y 2 ) 2 . . . . . .
Differentiating by first keeping y constant and then x
d x d k = 2 x − 2 x 1 + 2 x − 2 x 2 + . . . + 2 x − 2 x n
and
d y d k = 2 y − 2 y 1 + 2 y − 2 y 2 . . . . . . . + 2 y − 2 y n
equating
d x d k = 0 and d y d k = 0
we get x = n x 1 + x 2 + . . . . x n
y = n y 1 + y 2 + . . . . y n
Using second derivative test
k is minimum for the values of x and y .
Thus, minimum value of k is sum of squares of distances from c e n t r o i d
(denoted as G )
G ( 5 1 + 2 + 3 + 4 + 5 , 5 1 2 + 2 2 + 3 2 + 4 2 + 5 2 )
G ( 3 , 1 1 )
k = 2 2 + 1 0 2 + 1 2 + 7 2 + 0 2 + 2 2 + 1 2 + 5 2 + 2 2 + 1 4 2
k = 3 8 4
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