Hẫy chứng minh các tam giác ABC vuông biết: a) A(2; 5) B(1;2 ) C( 4:1 ) b) A(3:5 ) B(-1; 3) C( 0;1 )

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