a: \(AB=\sqrt{\left(1-3\right)^2+\left(5-1\right)^2}=\sqrt{2^2+4^2}=2\sqrt{5}\)
\(BC=\sqrt{\left(12-1\right)^2+\left(7-5\right)^2}=5\sqrt{5}\)
\(AC=\sqrt{\left(12-3\right)^2+\left(7-1\right)^2}=3\sqrt{13}\)
Chu vi tam giác ABC là:
\(C_{ABC}=AB+AC+BC=3\sqrt{13}+7\sqrt{5}\)
Xét ΔABC có \(cosBAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)
=>\(cosBAC=\dfrac{20+117-125}{2\cdot2\sqrt{5}\cdot3\sqrt{13}}=\dfrac{12}{12\sqrt{65}}=\dfrac{1}{\sqrt{65}}\)
=>\(sinBAC=\sqrt{1-\left(\dfrac{1}{\sqrt{65}}\right)^2}=\sqrt{\dfrac{64}{65}}=\dfrac{8}{\sqrt{65}}\)
Diện tích tam giác ABC là:
\(S_{BAC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinBAC\)
\(=\dfrac{1}{2}\cdot\dfrac{8}{\sqrt{65}}\cdot2\sqrt{5}\cdot3\sqrt{13}=4\cdot2\cdot3=24\)
b: E(x;y); A(3;1); B(1;5); C(12;7)
\(\overrightarrow{EA}=\left(3-x;1-y\right);\overrightarrow{EB}=\left(1-x;5-y\right);\overrightarrow{EC}=\left(12-x;7-y\right)\)
\(\overrightarrow{EA}+3\cdot\overrightarrow{EB}=2\overrightarrow{EC}\)
=>\(\left\{{}\begin{matrix}3-x+3\left(1-x\right)=2\left(12-x\right)\\1-y+3\left(5-y\right)=2\left(7-y\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3-x+3-3x=24-2x\\1-y+15-3y=14-2y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-4x+6=-2x+24\\-4y+16=-2y+14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2x=24-6=18\\-2y=14-16=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-9\\y=1\end{matrix}\right.\)
vậy: E(-9;1)
