# 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.

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) = ( , )

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)+(y12+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+y32) - 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.

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