Question

If G be the centroid of a triangle ABC and P be any other point in the plane, prove that PA²+ PB² + PC² = GA² + GB² + GC² + 3GP².

Answer 1

Let ΔABC be any triangle whose coordinates are`A(x_1,y_1):B(x_2,y_2):C(x_3,y_3)`. Let P be the origin and G be the centroid of the triangle.

We have to prove that,  

`PA^2+PB^2+PC_2=GA^2+GB^2+GC^2+3GP^2`.......(1) 

We know that the co-ordinates of the centroid G of a triangle whose vertices are 

`(x_1,y_1), (x_2,y_2),(x_3,y_3)`is - 

`G((x_1+x_2+x_3)/3,(y_1+y_2+y_3)/3)` 

In genral, the distance between `A(x_1,y_1)`and `B(x_2,y_2)` is given by,

`AB=sqrt((x_2-x_1)^2+(y_2-y_1)^2` 

So, 

`PA^2=(x_1-0)^2+(y_1-0)^2` 

`=x_1^2+y_1^2` 

`PB^2=(x_2-0)^2+(y_2-0)^2`  

`=x_2^2+y_2^2` 

`PC^2=(x_3-0)^2 +(y_3-0)^2` 

`=x_3^2+y_3^2`

 NOW, 

`GP^2=((x_1+x_2+x_3)/3-0)^2+((y_1+y_2+y_3)/3-0)^2`

`=((x_1+x_2+x_3)/9)^2 +((y_1+y_2+y_3)/9)^2` 

`GA^2=(x_1(-x_1+x_2+x3)/3)^2 +(y_1-(y_1+y_2+y_3)/3)^2` 

`((2x_1-x_2-x_3)/9)^2 +((2y^1-y_2-y_3)/9)^2`

`GB_2=(x^2-(x_1+x_2+x_3)/3)^2 + (y_2-(y_1+y_2+y_3)/3)^2`  

`((2x_2-x_1-x_3)/9)^2 +((2y_2-y_1-y_3)/9)^2` 

`GC^2=((x_3-x_1+x_2+x_3)/3)^2 +(y_3-(y_1+y_2+y_3)/3)^2`

`((2x_3-x_1-x_2)/9)^2 +((2y_3-y_1-y_-2)/9)^2` 

So we get the value of left hand side of equation (1) as 

`PA^2+PB^2+PC^2=x_1^2+x_2^2+x_3^2+y_1^2+y_2^2+y_3^2`

 Similarly we get the value of right hand side of equation (1) as 

`GA^2+GB^2+GC^2+3GP^2=[((2x_1-x_2-x_3)^2)/9] +[ (2x_2-x_2-x_3)^2/9+(2y_2-y_1-y_3)^2/9]` 

`+[(2x_3-x_1-x_2)^2/9+(2y_3-x_1-x_2)^2/9]+3[(x_1+x_2+x_3)^2/9+(y_1+y_2+y_3)^2/9+(y_1+y_2+y_3)^2/9]` 

`=[2/3(x_1^3+x_2^2+x_3^2)+1/3(x_1^2+x_2^2+x_3^2)] +[2/3(y_1^2+y_2^2+y_3^2)+1/3(y_1^2+y_2^2+y_3^2)]` 

`=x_1^2+x_2^2+x_3^2+y_1^2+y_2^2+y_2^2` 

Hence, 

`PA^2+PB^2+PC^2=GB^2+GC^2+3GP^2`