a: A(-1;2); B(3;4); C(-2;3)
Tọa độ vecto AB là:
\(\left\{{}\begin{matrix}x=3-\left(-1\right)=4\\y=4-2=2\end{matrix}\right.\)
Tọa độ vecto AC là:
\(\left\{{}\begin{matrix}x=-2-\left(-1\right)=-2+1=-1\\y=3-2=1\end{matrix}\right.\)
Tọa độ vecto BC là:
\(\left\{{}\begin{matrix}x=-2-3=-5\\y=3-4=-1\end{matrix}\right.\)
b: Tọa độ trọng tâm của ΔABC là:
\(\left\{{}\begin{matrix}x=\dfrac{-1+3+\left(-2\right)}{3}=\dfrac{3-3}{3}=0\\y=\dfrac{2+4+3}{3}=3\end{matrix}\right.\)
c: ABCD là hình bình hành
=>\(\overrightarrow{AB}=\overrightarrow{DC}\)
mà \(\overrightarrow{AB}=\left(4;2\right);\overrightarrow{DC}=\left(-2-x;3-y\right)\)
nên \(\left\{{}\begin{matrix}-2-x=4\\3-y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-2-4=-6\\y=3-2=1\end{matrix}\right.\)
Vậy: D(-6;1)
d: A(-1;2); B(3;4); C(-2;3); I(x;y)
\(\overrightarrow{IA}=\left(-1-x;2-y\right)\)
\(\overrightarrow{IB}=\left(3-x;4-y\right)\)
\(\overrightarrow{IC}=\left(-2-x;3-y\right)\)
\(2\overrightarrow{IA}+3\overrightarrow{IB}-4\overrightarrow{IC}=\overrightarrow{0}\)
=>\(\left\{{}\begin{matrix}2\left(-1-x\right)+3\left(3-x\right)-4\left(-2-x\right)=0\\2\left(2-y\right)+3\left(4-y\right)-4\left(3-y\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2-2x+9-3x+8+4x=0\\4-2y+12-3y-12+4y=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-x+15=0\\-y+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=15\\y=4\end{matrix}\right.\)
Vậy: I(15;4)
