Câu hỏi
Cho tam giác ABC có \(C\left( { - 1; - 1} \right);\,AB = \sqrt 5 .\) Phương trình đường thẳng AB : \(x + 2y - 3 = 0\). Trọng tâm \(G \in \left( \Delta \right):\,\,x + y - 2 = 0.\) Tìm A, B.
- A \(\left[ \begin{array}{l}
A\left( {4; \frac{1}{2}} \right);\,\,B\left( {6; - \frac{3}{2}} \right)\\
A\left( {6; - \frac{3}{2}} \right);\,\,B\left( {4; \frac{1}{2}} \right)
\end{array} \right.\) - B \(\left[ \begin{array}{l}
A\left( {4; - \frac{1}{2}} \right);\,\,B\left( {6; \frac{3}{2}} \right)\\
A\left( {6; \frac{3}{2}} \right);\,\,B\left( {4; - \frac{1}{2}} \right)
\end{array} \right.\) - C \(\left[ \begin{array}{l}
A\left( {4; - \frac{1}{2}} \right);\,\,B\left( {6; - \frac{3}{2}} \right)\\
A\left( {6; - \frac{3}{2}} \right);\,\,B\left( {4; - \frac{1}{2}} \right)
\end{array} \right.\) - D \(\left[ \begin{array}{l}
A\left( {-4; - \frac{1}{2}} \right);\,\,B\left( {6; - \frac{3}{2}} \right)\\
A\left( {6; - \frac{3}{2}} \right);\,\,B\left( {-4; - \frac{1}{2}} \right)
\end{array} \right.\)
Phương pháp giải:
Bước 1 : Giả sử \(\left\{ \begin{array}{l}A\left( {a;b} \right)\\B\left( {c;d} \right)\end{array} \right. \in AB \Rightarrow \left\{ \begin{array}{l}a + 2b - 3 = 0\\c + 2d - 3 = 0\end{array} \right.\)
\( \Rightarrow \left\{ \begin{array}{l}a = 3 - 2b\\c = 3 - 2d\end{array} \right. \Rightarrow \left\{ \begin{array}{l}A\left( {3 - 2b;b} \right)\\B\left( {3 - 2d;d} \right)\end{array} \right.\)
Bước 2 : Lập 2 phương trình :
Trọng tâm \(G\left( {\frac{{5 - 2b - 2d}}{3};\frac{{b + d - 1}}{3}} \right)\).
\(G \in \left( \Delta \right) \Rightarrow b + d + 2 = 0\,\,\left( 1 \right)\)
\(AB = \sqrt 5 \Rightarrow {\left( {2b - 2d} \right)^2} + {\left( {d - b} \right)^2} = 5\,\,\left( 2 \right)\)
Giải hệ \(\left\{ \begin{array}{l}\left( 1 \right)\\\left( 2 \right)\end{array} \right. \Rightarrow \left[ \begin{array}{l}A\left( {4; - \frac{1}{2}} \right);\,\,B\left( {6; - \frac{3}{2}} \right)\\A\left( {6; - \frac{3}{2}} \right);\,\,B\left( {4; - \frac{1}{2}} \right)\end{array} \right.\) .
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