Gọi \(D\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\left(4;4\right)\\\overrightarrow{DC}=\left(4-x;-1-y\right)\end{matrix}\right.\)
Do \(\overrightarrow{AB}=\overrightarrow{DC}\Rightarrow\left\{{}\begin{matrix}4-x=4\\-1-y=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\\y=-5\end{matrix}\right.\) \(\Rightarrow D\left(0;-5\right)\)
b/ Gọi pt AB có dạng \(y=ax+b\Rightarrow\left\{{}\begin{matrix}-a+b=-2\\3a+b=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=-1\end{matrix}\right.\) \(\Rightarrow y=x-1\)
Giao với Ox: \(y=0\Rightarrow x=1\Rightarrow\left(1;0\right)\)
c/ Của đường thẳng y=2 với cái gì bạn?