Đáp án:
$\begin{array}{l}
A\left( {2;4} \right);B\left( {1;2} \right);C\left( {6;2} \right)\\
a)\overrightarrow {AB} = \left( { - 1; - 2} \right);\overrightarrow {AC} = \left( {4; - 2} \right)\\
\Rightarrow \overrightarrow {AB} .\overrightarrow {AC} = \left( { - 1} \right).4 + \left( { - 2} \right).\left( { - 2} \right) = 0\\
\Rightarrow AB \bot AC\\
\Rightarrow \Delta ABC \bot A\\
b)\overrightarrow {AB} = \left( { - 1; - 2} \right);\overrightarrow {AC} = \left( {4; - 1} \right)\\
\Rightarrow AB = \sqrt {1 + {2^2}} = \sqrt 5 \\
AC = \sqrt {{4^2} + 1} = \sqrt {17} \\
\Rightarrow {S_{ABC}} = \frac{1}{2}.AC.AC = \frac{1}{2}.\sqrt 5 .\sqrt {17} = \frac{{\sqrt {35} }}{2}\\
c){y_B} = 2;{y_C} = 2\\
\Rightarrow BC:y = 2\\
\Rightarrow H\left( {x;2} \right)\\
\Rightarrow \overrightarrow {AH} = \left( {x - 2; - 2} \right)\\
AH \bot BC\\
\Rightarrow \overrightarrow {AH} .\overrightarrow {BC} = 0\\
\Rightarrow 5.\left( {x - 2} \right) - 2.0 = 0\\
\Rightarrow x = 2\\
Vậy\,H\left( {2;2} \right)
\end{array}$
