Đáp án:
$\begin{array}{l}
a)\overrightarrow {AB} = \left( {4; - 5} \right);\overrightarrow {AC} = \left( {5;4} \right)\\
\Rightarrow \overrightarrow {AB} .\overrightarrow {AC} = 4.5 - 5.4 = 0\\
b)\\
Do:\overrightarrow {AB} .\overrightarrow {AC} = 0\\
\Rightarrow AB \bot AC\\
\Rightarrow \Delta ABC\,vuông\,tại\,A
\end{array}$
c)
Gọi H(x;y)
$\begin{array}{l}
\Rightarrow \overrightarrow {BH} = \left( {x - 5;y + 2} \right);\overrightarrow {CH} = \left( {x - 6;y - 7} \right)\\
DO:\left\{ \begin{array}{l}
BH \bot AC\\
CH \bot AB
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\overrightarrow {BH} .\overrightarrow {AC} = 0\\
\overrightarrow {CH} .\overrightarrow {AB} = 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
5.\left( {x - 5} \right) + 4\left( {y + 2} \right) = 0\\
4.\left( {x - 6} \right) - 5\left( {y - 7} \right) = 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = 1\\
y = 3
\end{array} \right.\\
\Rightarrow H\left( {1;3} \right)\\
d)\\
\left\{ \begin{array}{l}
{x_I} = \frac{{{x_A} + {x_B}}}{2} = \frac{{1 + 5}}{2} = 3\\
{y_I} = \frac{{{y_A} + {y_B}}}{2} = \frac{{3 - 2}}{2} = \frac{1}{2}
\end{array} \right. \Rightarrow I\left( {3;\frac{1}{2}} \right)\\
\left\{ \begin{array}{l}
{x_G} = \frac{{{x_A} + {x_B} + {x_C}}}{3} = 4\\
{y_G} = \frac{{{y_A} + {y_B} + {y_C}}}{3} = \frac{8}{3}
\end{array} \right. \Rightarrow G\left( {4;\frac{8}{3}} \right)
\end{array}$
