Trong mặt phẳng  $Oxy$  cho hai điểm  $A(1;3),B(13;8).$
a) Xác định tọa độ  $\overrightarrow{AB}. $
b) Tìm khoảng cách giữa  $A$  và  $B$.
c) Tìm tọa độ $I$  là trung điểm của $AB$.
d) Tìm tọa độ  $A'$, điểm đối xứng của  $A$  qua  $B$.
a) Ta có:  $\overrightarrow{AB}=(13-1;8-3)    \Rightarrow      \overrightarrow{AB}=(12;5). $

b) Ta có:  $AB=|\overrightarrow{AB} |=\sqrt{12^2+5^2}=\sqrt{169}=13.  $

c) Vì  $I$ là trung điểm của  $AB$.
Ta có:  $\begin{cases}x_I=\frac{x_A+x_B}{2}  \\ y_I=\frac{y_A+y_B}{2}  \end{cases}    \Leftrightarrow      \begin{cases}x_I=\frac{1+13}{2}=7  \\ y_I=\frac{3+8}{2}=\frac{11}{2}   \end{cases}     \Rightarrow     I=(7; \frac{11}{2} )$.

d) Vì  $A'$  đối xứng với  $A$  qua  $B$, nên  $B$  là trung điểm  của  đoạn  $AA'$.
$\Leftrightarrow    \begin{cases}x_B=\frac{x_A+x_A'}{2}  \\ y_B=\frac{y_A+y_A'}{2}  \end{cases}       \Leftrightarrow    \begin{cases}x_A'=2x_B-x_A=26-1=25 \\ y_A'=2y_B-y_A=16-3=13 \end{cases} $
Vậy  $A'(25;13).$

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