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)

Answer 1

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)