Giải thích các bước giải:
Gọi $I$ là trung điểm của $AB$
Ta có:
$\begin{array}{l}
1)A\left( {2;3} \right),B\left( {1;4} \right)\\
\Rightarrow AB = \sqrt {{{\left( {{x_B} - {x_A}} \right)}^2} + {{\left( {{y_B} - {y_A}} \right)}^2}} = \sqrt {{{\left( {1 - 2} \right)}^2} + {{\left( {4 - 3} \right)}^2}} = \sqrt 2 \\
\Rightarrow I\left( {\dfrac{{{x_A} + {x_B}}}{2};\dfrac{{{y_A} + {y_B}}}{2}} \right) \Rightarrow I\left( {\dfrac{3}{2};\dfrac{7}{2}} \right)
\end{array}$
Vậy $AB = \sqrt 2 ;I\left( {\dfrac{3}{2};\dfrac{7}{2}} \right)$
$\begin{array}{l}
2)A\left( { - 3;2} \right),B\left( {1; - 5} \right)\\
\Rightarrow AB = \sqrt {{{\left( {{x_B} - {x_A}} \right)}^2} + {{\left( {{y_B} - {y_A}} \right)}^2}} = \sqrt {{{\left( {1 - \left( { - 3} \right)} \right)}^2} + {{\left( { - 5 - 2} \right)}^2}} = \sqrt {65} \\
\Rightarrow I\left( {\dfrac{{{x_A} + {x_B}}}{2};\dfrac{{{y_A} + {y_B}}}{2}} \right) \Rightarrow I\left( { - 1;\dfrac{{ - 3}}{2}} \right)
\end{array}$
Vậy $AB = \sqrt {65} ;I\left( { - 1;\dfrac{{ - 3}}{2}} \right)$
$\begin{array}{l}
3)A\left( { - 5; - 1} \right),B\left( {10; - 2} \right)\\
\Rightarrow AB = \sqrt {{{\left( {{x_B} - {x_A}} \right)}^2} + {{\left( {{y_B} - {y_A}} \right)}^2}} = \sqrt {{{\left( {10 - \left( { - 5} \right)} \right)}^2} + {{\left( { - 2 - \left( { - 1} \right)} \right)}^2}} = \sqrt {226} \\
\Rightarrow I\left( {\dfrac{{{x_A} + {x_B}}}{2};\dfrac{{{y_A} + {y_B}}}{2}} \right) \Rightarrow I\left( {\dfrac{5}{2};\dfrac{{ - 3}}{2}} \right)
\end{array}$
Vậy $AB = \sqrt {226} ;I\left( {\dfrac{5}{2};\dfrac{{ - 3}}{2}} \right)$
