\(\overrightarrow{AB}=\left(2;1\right)\)
\(\overrightarrow{AD}=\left(x-1;y+1\right)\)
Vì ABCDlà hình vuông
nên vecto AB*vecto AD=0 và AB=AD
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)+y+1=0\\2^2+1^2=\left(x-1\right)^2+\left(y+1\right)^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-2+y+1=0\\\left(x-1\right)^2+\left(y+1\right)^2=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y-1=0\\\left(x-1\right)^2+\left(y+1\right)^2=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-2x+1\\\left(x-1\right)^2+\left(-2x+1+1\right)^2=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-2x+1\\\left(x-1\right)^2+4\left(x-1\right)^2=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2x+1\\\left(x-1\right)^2=1\end{matrix}\right.\)
(x-1)^2=1
=>x-1=1 hoặc x-1=-1
=>x=0 hoặc x=2
Khi x=0 thì y=-2*0+1=1(loại)
Khi x=2 thì y=-2*2+1=-3(nhận)
Vậy: D(2;-3)
