Giải thích các bước giải:
a,
G là trọng tâm tam giác ABC nên:
\(\left\{ \begin{array}{l}
{x_G} = \frac{{{x_A} + {x_B} + {x_C}}}{3}\\
{y_G} = \frac{{{y_A} + {y_B} + {y_C}}}{3}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{x_G} = \frac{{1 + 2 + 1}}{3}\\
{y_G} = \frac{{4 - 3 - 2}}{3}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{x_G} = \frac{4}{3}\\
{y_G} = - \frac{1}{3}
\end{array} \right. \Rightarrow G\left( {\frac{4}{3}; - \frac{1}{3}} \right)\)
b,
Ta có:
\(\begin{array}{l}
A\left( {1;4} \right);\,\,\,B\left( {2; - 3} \right);\,\,C\left( {1; - 2} \right)\\
\overrightarrow {AB} \left( {1; - 7} \right) \Rightarrow AB = \sqrt {{1^2} + {{\left( { - 7} \right)}^2}} = 5\sqrt 2 \\
\overrightarrow {BC} \left( { - 1;1} \right) \Rightarrow BC = \sqrt {{{\left( { - 1} \right)}^2} + {1^2}} = \sqrt 2 \\
\overrightarrow {CA} \left( {0;6} \right) \Rightarrow CA = \sqrt {{0^2} + {6^2}} = 6\\
p = \frac{{AB + BC + CA}}{2} = 3\sqrt 2 + 3\\
{S_{ABC}} = \sqrt {p\left( {p - AB} \right)\left( {p - BC} \right)\left( {p - CA} \right)} = 3
\end{array}\)
