a; A(-1;5); B(5;-3)
=>\(\overrightarrow{AB}=\left(5+1;-3-5\right)\)
=>\(\overrightarrow{AB}=\left(6;-8\right)\)
\(AB=\sqrt{6^2+\left(-8\right)^2}=10\)
b: Tọa độ trung điểm I của AB là:
\(\left\{{}\begin{matrix}x_I=\dfrac{x_A+x_B}{2}=\dfrac{-1+5}{2}=\dfrac{4}{2}=2\\y_I=\dfrac{y_A+y_B}{2}=\dfrac{5-3}{2}=\dfrac{2}{2}=1\end{matrix}\right.\)
=>I(2;1)
G là trọng tâm của ΔABC
=>\(\left\{{}\begin{matrix}x_A+x_B+x_C=3\cdot x_G\\y_A+y_B+y_C=3\cdot y_G\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x_C-1+5=3\cdot\left(-2\right)=-6\\y_C+5-3=3\cdot4=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_C=-6+1-5=-5-5=-10\\y_C=12-2=10\end{matrix}\right.\)
=>C(-10;10)