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Question
Let A(-a, 0), B(0, a) and C(α , β) be the vertices of the L1 ABC and G be its centroid . Prove that
GA2 + GB2 + GC2 = `1/3` (AB2 + BC2 + CA2)
Solution
Coordinates of G are ,
G (x , y) = G `((-"a" + 0 + "a")/3 , (0 + "a" + "b")/3)` = G `(0 , ("a + b")/3)`
GA2 = (0 + a)2 + `(("a + b")/3 - 0)^2`
GA2 = `(9"a"^2 + "a"^2 + "b"^2 + 2"ab")/9 = (10"a"^2 + "b"^2 + 2"ab")/9`
GB2 = (0 - 0)2 + `(("a + b")/3 - "a")^2`
GB2 = `(("b" - 2"a")/3)^2 = ("b"^2 + 4"a"^2 - 4"ab")/9`
GC2 = (0 - a)2 + `(("a + b")/3 - "b")^2`
GC2 = a2 + `(("a - 2b")/3)^2 = (9"a"^2 + "a"^2 + 4"b"^2 - 4"ab")/9`
GA2 + GB2 + GC2 = `(10"a"^2 + "b"^2 + 2"ab" + "b"^2 + 4"a"^2 - 4"ab" + 10"a"^2 + 4"b"^2 - 4"ab")/9 `
= `(24"a"^2 + 6 "b"^2 - 6"ab")/9`
GA2 + GB2 + GC2 = `1/3` (8a2 + 2b2 - 2ab) .....(1)
AB2 = (- a - 0)2 + (0 - a)2 = 2a2
BC2 = (0 - a)2 + (a - b)2 = a2 + a2 + b2 - 2ab = 2a2 + b2 - 2ab
AC2 = (- a - a)2 + (0 - b)2 = 4a2 + b2
from (1) and (2)
GA2 + GB2 + GC2 = `1/3` (AB2 + BC2 + CA2)
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