a: A(3;2); B(1;-3); C(1;4)
Tọa độ vecto AB là:
\(\left\{{}\begin{matrix}x=x_B-x_A=1-3=-2\\y=y_B-y_A=-3-2=-5\end{matrix}\right.\)
Vậy: \(\overrightarrow{AB}=\left(-2;-5\right)\)
Tọa độ vecto AC là:
\(\left\{{}\begin{matrix}x=x_C-x_A=1-3=-2\\y=y_C-y_A=4-2=2\end{matrix}\right.\)
Vậy: \(\overrightarrow{AC}=\left(-2;2\right)\)
=>\(\overrightarrow{CA}=\left(2;-2\right)\)
Tọa độ vecto BC là:
\(\left\{{}\begin{matrix}x=x_C-x_B=1-1=0\\y=y_C-y_B=4-\left(-3\right)=7\end{matrix}\right.\)
Vậy: \(\overrightarrow{BC}=\left(0;7\right)\)
b: \(\overrightarrow{AB}=\left(-2;-5\right);\overrightarrow{AC}=\left(-2;2\right);\overrightarrow{BC}=\left(0;7\right)\)
\(AB=\sqrt{\left(-2\right)^2+\left(-5\right)^2}=\sqrt{29}\)
\(AC=\sqrt{\left(-2\right)^2+2^2}=2\sqrt{2}\)
\(BC=\sqrt{0^2+7^2}=7\)
Chu vi tam giác ABC là:
\(AB+AC+BC=2\sqrt{2}+\sqrt{29}+7\)
