Đặt cạnh hình vuông là x
\(MD=\sqrt{x^2+\left(\frac{x}{2}\right)^2}=\frac{x\sqrt{5}}{2}\)
\(JD=\sqrt{\left(\frac{x}{4}\right)^2+\left(\frac{3x}{4}\right)^2}=\frac{x\sqrt{10}}{4}\)
\(JM=\sqrt{\left(\frac{x}{4}\right)^2+\left(\frac{3x}{4}\right)^2}=\frac{x\sqrt{10}}{4}\)
\(\Rightarrow JD^2+JM^2=MD^2\Rightarrow\Delta JMD\) vuông cân tại J \(\Rightarrow DJ\perp MJ\)
\(\overrightarrow{MJ}=\left(1;-3\right)\Rightarrow\) phương trình JD:
\(1\left(x-1\right)-3y=0\Leftrightarrow x-3y-1=0\)
Tọa độ D là nghiệm: \(\left\{{}\begin{matrix}x-y+1=0\\x-3y-1=0\end{matrix}\right.\) \(\Rightarrow D\left(-2;-1\right)\)
\(JM=\sqrt{10}=\frac{x\sqrt{10}}{4}\Rightarrow x=4\)
Gọi \(A\left(a;b\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{DA}=\left(a+2;b+1\right)\\\overrightarrow{MA}=\left(a;b-3\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}AM\perp AD\\AD=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a\left(a+2\right)+\left(b-3\right)\left(b+1\right)=0\\\left(a+2\right)^2+\left(b+1\right)^2=16\end{matrix}\right.\)
\(\Rightarrow\) Tọa độ A
\(\Rightarrow\) Tọa độ B (M là trung điểm AB)
\(\Rightarrow\) Tọa độ C (\(\overrightarrow{AB}=\overrightarrow{DC}\))
