\[\begin{array}{l}
a)\,\,A\left( {2;\,5} \right);\,\,B\left( {1;\,\,2} \right);\,\,C\left( {4;\,\,1} \right)\\
\Rightarrow \overrightarrow {AB} = \left( { - 1;\,\, - 3} \right);\,\,\,\overrightarrow {BC} = \left( {3;\,\, - 1} \right)\\
\Rightarrow \overrightarrow {AB} .\overrightarrow {BC} = 3.\left( { - 1} \right) + \left( { - 3} \right)\left( { - 1} \right) = - 3 + 3 = 0\\
\Rightarrow \overrightarrow {AB} \bot \overrightarrow {BC} \Rightarrow \Delta ABC\,\,\,vuong\,\,tai\,\,B.\\
b)\,\,\,A\left( {3;\,\,5} \right);\,\,\,B\left( { - 1;\,\,3} \right);\,\,\,C\left( {0;\,\,1} \right)\\
\Rightarrow \overrightarrow {AB} = \left( { - 4;\,\, - 2} \right);\,\,\,\,\overrightarrow {AC} = \left( { - 3;\,\, - 4} \right);\,\,\overrightarrow {BC} = \left( {1;\,\, - 2} \right)\\
\Rightarrow \overrightarrow {AB} .\overrightarrow {BC} = - 4.1 + \left( { - 2} \right)\left( { - 2} \right) = - 4 + 4 = 0\\
\Rightarrow \overrightarrow {AB} \bot \overrightarrow {BC} \Rightarrow \Delta ABC\,\,\,vuong\,\,tai\,\,B.
\end{array}\]
