a:
Tọa độ vecto AB là:
\(\left\{{}\begin{matrix}x_{\overrightarrow{AB}}=x_B-x_A=-1-5=-6\\y_{\overrightarrow{AB}}=y_B-y_A=-3-2=-5\end{matrix}\right.\)
=>\(\overrightarrow{AB}=\left(-6;-5\right)\)
Tọa độ vecto AC là: \(\left\{{}\begin{matrix}x_{\overrightarrow{AC}}=x_C-x_A=2-5=-3\\y_{\overrightarrow{AC}}=y_C-y_A=4-2=2\end{matrix}\right.\)
=>\(\overrightarrow{AC}=\left(-3;2\right)\)
Tọa độ vecto BC là:
\(\left\{{}\begin{matrix}x_{\overrightarrow{BC}}=x_C-x_B=2-\left(-1\right)=3\\y_{\overrightarrow{BC}}=y_C-y_B=4-\left(-3\right)=4+3=7\end{matrix}\right.\)
=>\(\overrightarrow{BC}=\left(3;7\right)\)
b: \(AB=\sqrt{\left(-1-5\right)^2+\left(-3-2\right)^2}=\sqrt{6^2+5^2}=\sqrt{61}\)
\(AC=\sqrt{\left(2-5\right)^2+\left(4-2\right)^2}=\sqrt{\left(-3\right)^2+2^2}=\sqrt{13}\)
\(BC=\sqrt{\left(2+1\right)^2+\left(4+3\right)^2}=\sqrt{3^2+7^2}=\sqrt{58}\)
c: Tọa độ trung điểm I của AB là:
\(\left\{{}\begin{matrix}x_I=\dfrac{x_A+x_B}{2}=\dfrac{5+\left(-1\right)}{2}=\dfrac{4}{2}=2\\y_I=\dfrac{2+\left(-3\right)}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
=>I(2;-1/2)
Tọa độ trọng tâm G của ΔABC là:
\(\left\{{}\begin{matrix}x_G=\dfrac{5+\left(-1\right)+2}{3}=\dfrac{7-1}{3}=\dfrac{6}{3}=2\\y_G=\dfrac{2+\left(-3\right)+4}{3}=\dfrac{6-3}{3}=\dfrac{3}{3}=1\end{matrix}\right.\)
=>G(2;1)
d: ABCD là hình bình hành
=>\(\overrightarrow{AB}=\overrightarrow{DC}\)
mà \(\overrightarrow{AB}=\left(-6;-5\right);\overrightarrow{DC}=\left(2-x;4-y\right)\)
nên \(\left\{{}\begin{matrix}2-x=-6\\4-y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=9\end{matrix}\right.\)
=>D(8;9)
