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Tọa độ trọng tâm G x G ; y G là x G = 1 − 2 + 5 3 = 4 3 y G = 3 + 4 + 3 3 = 10 3 .
Chọn D.
Gọi C(x;y) \(\Rightarrow\left\{{}\begin{matrix}x_G=\dfrac{x+2}{3}\\y_G=\dfrac{y-6}{3}\end{matrix}\right.\) \(\Rightarrow3\left(\dfrac{x+2}{3}\right)-\dfrac{y-6}{3}+1=0\)
\(\Leftrightarrow3x-y+15=0\Rightarrow y=3x+15\Rightarrow C\left(x;3x+15\right)\)
\(S_{ABC}=\dfrac{1}{2}\left|\left(x_B-x_A\right)\left(y_C-y_A\right)-\left(x_C-x_A\right)\left(y_B-y_A\right)\right|\)
\(\Leftrightarrow3=\dfrac{1}{2}\left|-2\left(3x+19\right)-2\left(x-2\right)\right|\)
\(\Rightarrow x=...\)
\(\overrightarrow{GM}=\left(-\dfrac{1}{3};0\right)\)
Gọi \(A\left(x;y\right)\Rightarrow\overrightarrow{AM}=\left(1-x;1-y\right)\)
\(\overrightarrow{AM}=3\overrightarrow{GM}\Rightarrow\left\{{}\begin{matrix}1-x=-1\\1-y=0\end{matrix}\right.\) \(\Rightarrow A\left(2;1\right)\)
DO B thuộc x+y-7=0 \(\Rightarrow B\left(x;7-x\right)\)
\(\left\{{}\begin{matrix}x_C=3x_G-x_A-x_B=2-x\\y_C=3y_G-y_A-y_B=x-5\end{matrix}\right.\) \(\Rightarrow C\left(2-x;x-5\right)\)
\(\Rightarrow\overrightarrow{AC}=\left(-x;x-6\right)\)
Do AC vuông góc x+y-7=0 \(\Rightarrow\dfrac{-x}{1}=\dfrac{x-6}{1}\Rightarrow x=3\Rightarrow\left\{{}\begin{matrix}B\left(3;4\right)\\C\left(-1;-2\right)\end{matrix}\right.\)