\(\overrightarrow{AB}=\left(4;-3\right)\Rightarrow AB=5\)

\(\overrightarrow{AC}=\left(6;0\right)\Rightarrow AC=6\)

\(\overrightarrow{BC}=\left(2;3\right)\Rightarrow BC=\sqrt{13}\)

Chu vi tam giác: \(AB+AC+BC=11+\sqrt{13}\)

\(\overrightarrow{AB}=\left(1;-2\right)\Rightarrow AB=\sqrt{5}\)

\(\overrightarrow{AC}=\left(-2;2\right)\Rightarrow AC=2\sqrt{2}\)

\(BC=\left(-3;4\right)\Rightarrow BC=5\)

Chu vi tam giác ABC: \(AB+AC+BC=\sqrt{5}+2\sqrt{2}+5\)

\(AB=\sqrt{\left(0+1\right)^2+\left(2+3\right)^2}=\sqrt{26}\)

\(AC=\sqrt{\left(2+1\right)^2+\left(1+3\right)^2}=\sqrt{3^2+4^2}=5\)

\(BC=\sqrt{\left(2-0\right)^2+\left(1-2\right)^2}=\sqrt{5}\)

=>\(C=\sqrt{26}+5+\sqrt{5}\left(cm\right)\)

\(AB=\sqrt{\left(5-1\right)^2+\left(-3+1\right)^2}=2\sqrt{5}\)

\(AC=\sqrt{\left(0-1\right)^2+\left(1+1\right)^2}=\sqrt{5}\)

\(BC=\sqrt{\left(0-5\right)^2+\left(1+3\right)^2}=\sqrt{29}\)

=>C=3 căn 5+căn 29

Gọi trực tâm của ΔABC là H(x;y)

=>AH⊥BC và BH⊥AC

=>\(\overrightarrow{AH}\cdot\overrightarrow{BC}=0;\overrightarrow{BH}\cdot\overrightarrow{AC}=0\)

A(-4;1); B(-1;4); C(3;-2); H(x;y)

\(\overrightarrow{AH}=\left(x+4;y-1\right);\overrightarrow{BC}=\left(3+1;-2-4\right)=\left(4;-6\right)\)

\(\overrightarrow{AH}\cdot\overrightarrow{BC}=0\)

=>4(x+4)+(-6)(y-1)=0

=>4x+16-6y+6=0

=>4x-6y+22=0

=>2x-3y+11=0

\(\overrightarrow{BH}=\left(x+1;y-4\right);\overrightarrow{AC}=\left(3+4;-2-1\right)=\left(7;-3\right)\)

\(\overrightarrow{BH}\cdot\overrightarrow{AC}=0\)

=>7(x+1)+(-3)(y-4)=0

=>7x+7-3y+12=0

=>7x-3y+19=0

=>7x-3y+19-2x+3y-11=0

=>5x+8=0

=>5x=-8

=>x=-1,6

2x-3y+11=0

=>-3,2-3y+11=0

=>-3y+7,8=0

=>-3y=-7,8

=>y=2,6

Vậy: H(-1,6;2,6)

O 1 4 y -4 -1 1 A B C x 2

Tam giác ABC là tam giác vuông

AB=5, BC=6

diện tích tam giác ABC là 5.6:2=15 (dvdt)