Đáp án:
Giải thích các bước giải:
a) Ta có: $\vec{AC}$=(12;-2); $\vec{BC}$=(6;-6)
Gọi H là trực tâm $\triangle$ABC: H(x;y)
Có: $\vec{AH}$=(x;y-1); $\vec{BH}$=(x-6;y-5)
H là trực tâm $\triangle$ABC $\Leftrightarrow$ $$ \left\{\begin{matrix} \vec{AH}.\vec{BC}=0\\\vec{BH}.\vec{AC}=0 \end{matrix}\right.$$
$\Leftrightarrow$ $$ \left\{\begin{matrix} x.6+(y-1).(-6)=0\\(x-6).12+(y-5).(-2)=0 \end{matrix}\right.$$
$\Leftrightarrow$ $$ \left\{\begin{matrix} x-y=-1\\6x-y=31 \end{matrix}\right.$$
$\Leftrightarrow$ $$ \left\{\begin{matrix} x=\frac{32}{5}\\y=\frac{37}{5} \end{matrix}\right.$$
$\Rightarrow$ H($\frac{32}{5}$;$\frac{37}{5}$
b) Tọa độ trọng tâm G của $\triangle$ABC: $$ \left\{\begin{matrix} x=\frac{x_{A}+x_{B}+x_{c}}{3}\\y=\frac{y_{A}+y_{B}+y_{c}}{3} \end{matrix}\right.$$
$\Leftrightarrow$ $$ \left\{\begin{matrix} x=\frac{0+6+12}{3}\\y=\frac{1+5-1}{3} \end{matrix}\right.$$
$\Leftrightarrow$ $$ \left\{\begin{matrix} x=6\\y=\frac{5}{3} \end{matrix}\right.$$
