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 + 3 GP².

Answer 1

we will solve it by taking the coordinates A(x₁,y₁), B(x₂,y₂) and C(x₃,y₃)

 Let the co ordinates of the centroid be G(u, v). 

G(u, v) = \((\frac{x_1+x_2+x_3}3,\frac{y_1+y_2+y_3}3)\)

let the coordinates of P(h, k). 

now we will find L.H.S and R.H.S. separately. 

PA²+PB²+PC² 

= (h - x₁)² +(k - y₁)² +(h - x₂)²+ (k - y₂)² +(h - x₃)² +(k - y₃)² …by distance formula. 

= 3(h²+k²)+(x₁²+x₂²+x₃²)+(y₁²+y₂²+y₃²)-2h(x₁+x₂+x₃)-2k(y₁+y₂+y₃) 

= 3(h²+k²)+(x₁²+x₂²+x₃²)+(y1 2+y₂²+y₃²)-2h(3u)-2k(3v)

GA²+GB²+GC²+3GP² 

= (u-x₁)²+(v-y₁)²+(u-x₂)²+(v-y₂)²+(u-x₃)²+(v-y₃)²+3[(u-h)²+(v-k)²] …by distance formula. 

= 3(u²+v²)+(x₁²+y₁²+x₂²+y₂²+x₃²+y₂³) - 2u(x₁+x₂+x3) -2v(y₁+y₂+y₃) + 3[u²+h²-2uh+v²+k²-2vk] 

= 6(u²+v²)+(x₁²+y₁²+x₂²+y2²+x₃²+y₃²)-2u(3u)-2v(3v) +3(h²+k²) -6uh-6vk 

= (x₁²+x₂²+x₃²)+(y₁²+y₂²+y₃²)+3(h₂+k²)-6uh-6vk 

Hence LHS = RHS 

(The above relation is known as Leibniz Relation) 

Hence Proved.