# Bài 36 trang 31 SGK Hình học 10 Nâng cao

##### Hướng dẫn giải

a) Gọi $$G$$ là trọng tâm tam giác $$ABC$$, ta có

\eqalign{ & \left\{ \matrix{ {x_G} = {1 \over 3}({x_A} + {x_B} + {x_C}) = {1 \over 3}( - 4 + 2 + 2) = 0 \hfill \cr {y_G} = {1 \over 3}({y_A} + {y_B} + {y_C}) = {1 \over 3}(1 + 4 - 2) = 1 \hfill \cr} \right.\,\, \cr & \Rightarrow \,\,G\,(0\,;\,1). \cr}

b) Gọi $$D\,({x_{D\,}}\,;\,{y_D})$$  sao cho $$C$$ là trọng tâm tam giác $$ABD$$. Ta có

\eqalign{ & \left\{ \matrix{ {x_C} = {1 \over 3}({x_A} + {x_B} + {x_D}) \hfill \cr {y_C} = {1 \over 3}({y_A} + {y_B} + {y_D}) \hfill \cr} \right.\,\, \Rightarrow \left\{ \matrix{ 2 = {1 \over 3}( - 4 + 2 + {x_D}) \hfill \cr - 2 = {1 \over 3}(1 + 4 + {y_D}) \hfill \cr} \right. \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\left\{ \matrix{ {x_D} = 8 \hfill \cr {y_D} = - 11 \hfill \cr} \right. \cr & \Rightarrow \,\,D\,(8\,;\, - 11) \cr}

c) Gọi $$E({x_E}\,;\,{y_E})$$ sao cho $$ABCE$$ là hình bình hành. Ta có

\eqalign{ & \overrightarrow {AB} = \overrightarrow {EC} \,\,\,\, \Leftrightarrow \,\,(6\,;\,3) = (2 - {x_E}\,;\, - 2 - {y_E}) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\left\{ \matrix{ {x_E} = - 4 \hfill \cr {y_E} = - 5 \hfill \cr} \right. \cr & \Rightarrow \,\,E\,( - 4\,;\, - 5). \cr}