Question

A(-1,3) B(3,-1) si C(-1,1) , daca G este centrul de greutate al triunghiului ABC , arătați ca GA=GB

Answer 1

$r_{G} = \frac{r_{A}+r_{B}+r_{C}}{3} \\ r_{G}= \frac{(-i+3j)+(3i-j)+(-i-j)}{3} \\ r_{G} = \frac{i+j}{3} = \frac{1}{3}i+\frac{1}{3}j$

=> G($\frac{1}{3}, \frac{1}{3}$)

G(1/3,1/3);   A(-1,3);
$GA = \sqrt{ (-1- \frac{1}{3} )^{2} + (3-\frac{1}{3})^{2} } = \sqrt{ (-\frac{4}{3})^{2} +(\frac{8}{3})^{2} } = \sqrt{ \frac{16}{9}+\frac{64}{9} }= \frac{ \sqrt{80} }{3}$

G(1/3,1/3);   B(3,-1);
$GB=\sqrt{( 3 - \frac{1}{3} )^{2} + (-1- \frac{1}{3} )^{2} } = \sqrt{ (\frac{8}{3})^{2} +(- \frac{4}{3})^{2} } = \sqrt{ \frac{64}{9}+ \frac{16}{9} }= \frac{\sqrt{80} }{3}$

GA=GB.