Question

If $G$ be the centroid of a triangle $ABC$ and $P$ be any other point in the plane,then $P{A}^2+P{B}^2+P{C}^2=G{A}^2+G{B}^2+G{C}^2+4G{P}^2$.

• A. True
• B. False

Answer 1

The correct option is B False

Let the points be $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$

and The point P be (0,0,0)

Taking L.H.S.

$P{A}^2+P{B}^2+P{C}^2$ = $x_1^2+x_2^2+x_3^2+y_1^2+y_2^2+y_3^2-\left(1\right)$

Cordinates of centroid G=$(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

$G{A}^2=(\frac{x_2+x_3-2x_1}{3}{)}^2+(\frac{y_3+y_2-2y_1}{3}{)}^2-\left(2\right)$

$G{B}^2=(\frac{x_1+x_3-2x_2}{3}{)}^2+(\frac{y_1+y_3-2y_2}{3}{)}^2-\left(3\right)$

$G{C}^2=(\frac{x_1+x_2-2x_3}{3}{)}^2+(\frac{y_1+y_2-2y_3^2}{3}{)}^2-\left(4\right)$

$G{P}^2=(\frac{x_1+x_2+x_3}{3}{)}^2+(\frac{y_1+y_2+y_3}{3}{)}^2-\left(5\right)$

On Adding eqs.(2),(3),(4) and (5)

we get $G{A}^2+G{B}^2+G{C}^2+3G{P}^2$

=$x_1^2+x_2^2+x_3^2+y_1^2+y_2^2+y_3^2-\left(6\right)$

From eqs,(1) and (6)

L.H.S.=R.H.S

But in this case L.H.S is not equal to R.H.S