Question

Please solve question number 18 and 19 ASAP. Its important for my exam.

Answer 1

Q18.

The formula  GA² + GB² + GC² = 1/3 * (AB² + BC² + CA² ) is VALID for all triangles ABC.

Given A(-a,0),  B(0, a), C(α, β).

Let the Centroid G = (x , y) 
          x = (-a + 0 + α) /3 = (α - a)/3
          y = (0 + a + β) / 3 = (a + β)/3

GA² = (α +2a)²/3²  + (a+β)²/3²     =  [ α² + 5a²+4aα + β² + 2 aβ ] /9  
GB² = (α - a)²/3² + (β - 2a)²/3²     =  [ α² + 5a² - 2aα + β² - 4aβ ] / 9
GC² = (-a - 2α)²/3²  + (a - 2β)²/3²  = [ 2a² + 4α² + 4aα + 4β² -4aβ ] / 9

    LHS = (6 α² + 12 a² + 6 β² + 6 aα - 6 aβ) / 9

AB² = (0+a)²+ (a-0)² = 2a² 
BC² = α² + (β - a)²
CA² = (α+a)² + β²

    RHS = (4 a² + 2 α² + 2β² - 2 a β + 2 a α) /3

So you see that  LHS = RHS

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Q19

Given  ΔABC:   A(-3,2)   B(3,2)   C(-2,-3).

   => Centroid G :  [ (-3+3-2)/3 , (2+2-3)/3 ] = [ -2/3 , 1/3 ] 

Midpoint of line segment between (a, b) & (c , d) is :   [ (a+c)/2 , (b+d)/2]

     F = Midpoint of AB:   [ (-3+3)/2 , (2+2)/2 ] = [0, 2]
     E = Midpoint of CA :  [ (-3-2)/2 , (2-3)/2 ] =  [-5/2 , -1/2 ]
     D = Midpoint of BC :  [ (3-2)/2 , (2-3)/2 ] = [1/2 , -1/2 ]

     =>  Centroid of ΔDEF:  [ (0-5/2 +1/2 )/3 ,  (2 -1/2 -1/2)/3 ] =  [ 2/3, 1/3 ]

We can see that the coordinates of G are same in both cases.