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