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position vector of p = b +1/3 (b-c )
= 4/3b - 1/3c
position vector of q = c + 1/3 ( c-a )
= 4/3 c - 1/3 a
of r = 4/3 a - 1/3 b
eq. of line AP,
r = a + n1 ( 4/3b - 1/3c -a ) = (1-n1 )a + 4/3 n1 b - 1/3 n1 c ........................1
line BQ ,
r = b + n2 ( 4/3 c - 1/3 a -b ) = -1/3n2 a + (1- n2 )b + 4/3 n2 c ...............................2
line CR ,
r = c + n3 (4/3 a - 1/3 b - c ) = 4/3 n3 a -1/3 n3 b + (1-n3 )c .........................................3
position vector of X will be found by solving eq 1 &2 simultaneously , becoz it is the intersecting point of AP & BQ,
equating both eq.
(1-n1 )a + 4/3 n1 b - 1/3 n1 c = -1/3n2 a + (1- n2 )b + 4/3 n2 c
comparing coefficients of a, b & c
1-n1 = -1/3 n2 , 4/3 n1 = (1- n2 ) , - 1/3 n1 = 4/3 n2
gives n1 = 12/13, so the position vector of X will be, (putting n1 = 12/13 in eq 1 )
x = 1/13 a + 16/13 b -4/13 c .
in the similar way we can solve eq 2 &3 & get Y & solve 1 & 3 to get Z
Y = 1/13 b + 16/13 c - 4/13 a
Z = 1/13 c + 16/13 a - 4/13 b
centroid of the triangle XYZ
= (x + y +z )1/3
= (1/13 + 16/13 - 4/13 ) ( a+b +c ) /3
= ( a+b +c ) /3
, which is also the centroid of triangle ABC
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