Question

If

G

is the centroid of triangle

A B C

and

O

any other point, prove that:

3 (G A^{2} + G B^{2} + G C^{2}) = B C^{2} + C A^{2} + A B^{2}

Accepted Answer

Let, $A (x_{1}, y_{1}), B (x_{2}, y_{2})$ and $C (x_{3}, y_{3})$ be the vertices of $△ A B C$ .

Assume that the centroid of the $△ A B C$ be at the origin, i,e. $G = (0, 0)$

Centroid of the $△ A B C = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$

$∴ \frac{x_{1} + x_{2} + x_{3}}{3} = 0$ and $\frac{y_{1} + y_{2} + y_{3}}{3} = 0$

$\Rightarrow x_{1} + x_{2} + x_{3} = 0$ and $y_{1} + y_{2} + y_{3} = 0$

Squaring both sides, we get

${x_{1}}^{2} + {x_{2}}^{2} + {x_{3}}^{2} + 2 x_{1} x_{2} + 2 x_{2} x_{3} + 2 x_{3} x_{1} = 0$ and ${y_{1}}^{2} + {y_{2}}^{2} + {y_{3}}^{2} + 2 y_{1} y_{2} + 2 y_{2} y_{3} + 2 y_{3} y_{1} = 0$ ................. $(1)$

Now, RHS $= A B^{2} + B C^{2} + C A^{2}$

$= [{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}] + [{(x_{3} - x_{2})}^{2} + {(y_{3} - y_{2})}^{2}] + [{(x_{1} - x_{3})}^{2} + {(y_{1} - y_{3})}^{2}]$

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

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

$= (3 {x_{1}}^{2} + 3 {x_{2}}^{2} + 3 {x_{3}}^{2}) + (3 {y_{1}}^{2} + 3 {y_{2}}^{2} + 3 {y_{3}}^{2})$ (from $(1)$ )

$= 3 ({x_{1}}^{2} + {x_{2}}^{2} + {x_{3}}^{2}) + 3 ({y_{1}}^{2} + {y_{2}}^{2} + {y_{3}}^{2})$ ................ $(2)$

Now, LHS $= 3 (G A^{2} + G B^{2} + G C^{2})$

$= 3 [\{{(x_{1} - 0)}^{2} + {(y_{1} - 0)}^{2}\} + \{{(x_{2} - 0)}^{2} + {(y_{2} - 0)}^{2}\} + \{{(x_{3} - 0)}^{2} + {(y_{3} - 0)}^{2}\}]$

$= 3 [({x_{1}}^{2} + {y_{1}}^{2}) + ({x_{2}}^{2} + {y_{2}}^{2}) + ({x_{3}}^{2} + {y_{3}}^{2})]$

$= 3 ({x_{1}}^{2} + {x_{2}}^{2} + {x_{3}}^{2}) + 3 ({y_{1}}^{2} + {y_{2}}^{2} + {y_{3}}^{2})$ .................... $(3)$

From $(2)$ and $(3)$ , we get

$3 (G A^{2} + G B^{2} + G C^{2}) = A B^{2} + B C^{2} + C A^{2}$

Thus, LHS $=$ RHS

Hence proved.

H K Dass, Rama Verma and, Bhagwat Swarup Sharma Solutions for Chapter: Co-Ordinate Geometry, Exercise 1: EXERCISE

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