\(M\left(x;y\right)\in\left(d\right):y=2x-1\Leftrightarrow M\left(x;2x-1\right)\)
\(\overrightarrow{MA}=\left(x+2;2x-4\right)\)
\(\overrightarrow{MB}=\left(x-2;2x-5\right)\)
\(\overrightarrow{MC}=\left(x-3;2x-1\right)\)
\(\overrightarrow{MD}=\left(x+1;2x\right)\)
\(\Rightarrow\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}=\left(3x-3;6x-10\right)\)
\(\Rightarrow\left(\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right).\overrightarrow{MD}=\left(3x-3\right)\left(x+1\right)+\left(6x-10\right).2x=-3\)
\(\Leftrightarrow3x^2-3+12x^2-20x=-3\)
\(\Leftrightarrow15x^2-20x=0\)
\(\Leftrightarrow5x\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\Rightarrow y=-1\\x=\dfrac{4}{3}\Rightarrow y=\dfrac{5}{3}\end{matrix}\right.\)
Vậy có \(2\) điểm \(M\left(0;-1\right)\&M\left(\dfrac{4}{3};\dfrac{5}{3}\right)\) thỏa mãn đề bài
