Đáp án:
\(\begin{array}{l}
a)\,\,D\left( {2;\,\,9} \right)\\
b)\,\,E\left( {\frac{{11}}{4};\,\,2} \right)\\
c)\,\,F\left( {5;\,\,\frac{9}{2}} \right)
\end{array}\)
Giải thích các bước giải:
Câu 6:
\(A\left( {1;\, - 2} \right),\,\,\,B\left( {4;\,\,0} \right),\,\,C\left( { - 1;\,\,7} \right).\)
a) Gọi \(D\left( {a;\,\,b} \right)\)
Tứ giác ABDC là hình bình hành
\(\begin{array}{l} \Leftrightarrow \overrightarrow {AB} = \overrightarrow {CD} \Leftrightarrow \left( {3;\,\,2} \right) = \left( {a + 1;\,\,b - 7} \right)\\ \Leftrightarrow \left\{ \begin{array}{l}a + 1 = 3\\b - 7 = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 2\\b = 9\end{array} \right. \Rightarrow D\left( {2;\,\,9} \right).\end{array}\)
b) Gọi \(E\left( {m;\,n} \right)\)
\(\begin{array}{l} \Rightarrow \overrightarrow {AE} = - 3\overrightarrow {BE} \Leftrightarrow \left( {m - 1;\,\,n + 2} \right) = - 3\left( {m - 4;\,\,n} \right)\\ \Leftrightarrow \left\{ \begin{array}{l}m - 1 = - 3\left( {m - 4} \right)\\n + 2 = - 3n\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m - 1 = - 3m + 12\\4n = 2\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}4m = 11\\n = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m = \frac{{11}}{4}\\n = 2\end{array} \right. \Rightarrow E\left( {\frac{{11}}{4};\,\,2} \right).\end{array}\)
c) Gọi \(F\left( {x;\,\,y} \right)\)
\(\begin{array}{l} \Rightarrow \overrightarrow {AF} + 2\overrightarrow {BF} - \overrightarrow {CF} = \overrightarrow 0 \\ \Leftrightarrow \left( {x - 1;\,\,y + 2} \right) + 2\left( {x - 4;\,\,y} \right) - \left( {x + 1;\,\,y - 7} \right) = \left( {0;\,\,0} \right)\\ \Leftrightarrow \left\{ \begin{array}{l}x - 1 + 2\left( {x - 4} \right) - x - 1 = 0\\y + 2 + 2y - y + 7 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2x = 10\\2y = 9\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 5\\y = \frac{9}{2}\end{array} \right. \Rightarrow F\left( {5;\,\,\frac{9}{2}} \right).\end{array}\)
