Resposta :
[tex]A)\\x_g=\frac{2+(-4)+0}{3}=\frac{-2}{3}\\\\y_g=\frac{-6+2+4}{3}=0\\\\G=(-\frac{2}{3},0)\\\\B)\\A(2,-6)\\BC\rightarrow x_{BC}=\frac{-4}{2}=-2\rightarrow y_{BC}=\frac{6}{2}=3\\BC(-2,3)\\d_{A,BC}=\sqrt{4^2+9^2}=\sqrt{16+9}=\sqrt{25}=5\\\\d_{A,BC}=5[/tex]