Đáp án:
b) \(I\left( {\dfrac{5}{2};\dfrac{7}{2}} \right)\) và \(G\left( {\dfrac{{14}}{3};2} \right)\)
c) D(2;-4)
Giải thích các bước giải:
a) Ta có:
\(\overrightarrow {AB} = \left( {7;3} \right);\overrightarrow {AC} = \left( {10; - 3} \right);\overrightarrow {BC} = \left( {3; - 6} \right)\)
Suy ra \(\dfrac{7}{{10}} \ne - \dfrac{3}{3}\)
b) \(\begin{array}{l}
\left\{ \begin{array}{l}
{x_I} = \dfrac{{{x_A} + {x_B}}}{2} = \dfrac{5}{2}\\
{y_I} = \dfrac{{{y_A} + {y_B}}}{2} = \dfrac{7}{2}
\end{array} \right. \Rightarrow I\left( {\dfrac{5}{2};\dfrac{7}{2}} \right)\\
\left\{ \begin{array}{l}
{x_G} = \dfrac{{{x_A} + {x_B} + {x_C}}}{3} = \dfrac{{14}}{3}\\
{y_G} = \dfrac{{{y_A} + {y_B} + {y_C}}}{3} = 2
\end{array} \right. \Rightarrow G\left( {\dfrac{{14}}{3};2} \right)
\end{array}\)
c) \(\begin{array}{l}
ABCD\,la\,hbh\, \Rightarrow \overrightarrow {AB} = \overrightarrow {DC} \\
D\left( {x;y} \right) \Rightarrow \overrightarrow {DC} = \left( {9 - x; - 1 - y} \right)\\
\Rightarrow \left\{ \begin{array}{l}
9 - x = 7\\
- 1 - y = 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
y = - 4
\end{array} \right. \Rightarrow D\left( {2; - 4} \right)
\end{array}\)
