Q. 53.7( 10 Votes )

# If G be the centroid of a triangle ABC and P be any other point in the plane, prove that PA^{2} + PB^{2} + PC^{2} = GA^{2} + GB^{2} + GC^{2} + 3 GP^{2}.

Answer :

we will solve it by taking the coordinates A(x_{1},y_{1}), B(x_{2},y_{2}) and C(x_{3},y_{3})

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.

PA^{2}+PB^{2}+PC^{2}

= (h - x_{1})^{2} +(k - y_{1})^{2} +(h - x_{2})^{2}+ (k - y_{2})^{2} +(h - x_{3})^{2} +(k - y_{3})^{2} …by distance formula.

= 3(h^{2}+k^{2})+(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})+(y_{1}^{2}+y_{2}^{2}+y_{3}^{2})-2h(x_{1}+x_{2}+x_{3})-2k(y_{1}+y_{2}+y_{3})

= 3(h^{2}+k^{2})+(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})+(y_{1}^{2}+y_{2}^{2}+y_{3}^{2})-2h(3u)-2k(3v)

GA^{2}+GB^{2}+GC^{2}+3GP^{2}

= (u-x_{1})^{2}+(v-y_{1})^{2}+(u-x_{2})^{2}+(v-y_{2})^{2}+(u-x_{3})^{2}+(v-y_{3})^{2}+3[(u-h)^{2}+(v-k)^{2}] ……by distance formula.

= 3(u^{2}+v^{2})+(x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}+x_{3}^{2}+y_{3}^{2}) - 2u(x_{1}+x_{2}+x_{3}) - 2v(y_{1}+y_{2}+y_{3}) + 3[u^{2}+h^{2}-2uh+v^{2}+k^{2}-2vk]

= 6(u^{2}+v^{2})+(x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}+x_{3}^{2}+y_{3}^{2})-2u(3u)-2v(3v) +3(h^{2}+k^{2}) -6uh-6vk

= (x_{1}^{2}+x_{2}^{2}+x_{3}^{2})+(y_{1}^{2}+y_{2}^{2}+y_{3}^{2})+3(h^{2}+k^{2})-6uh-6vk

Hence LHS = RHS

(The above relation is known as Leibniz Relation)

Hence Proved.

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