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

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

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

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: \(AB+AC+BC=\sqrt{5}+2\sqrt{2}+5\)

Do P thuộc Ox nên tọa độ có dạng \(P\left(p;0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{MN}=\left(1;-3\right)\\\overrightarrow{MP}=\left(p-2;-1\right)\end{matrix}\right.\)

Do tam giác MNP vuông tại M \(\Rightarrow\overrightarrow{MN}.\overrightarrow{MP}=0\)

\(\Rightarrow1.\left(p-2\right)+3=0\) \(\Rightarrow p=-1\)

\(\Rightarrow P\left(-1;0\right)\)

\(\Rightarrow\overrightarrow{MP}=\left(-3;-1\right)\Rightarrow\left\{{}\begin{matrix}MN=\sqrt{1^2+\left(-3\right)^2}=\sqrt{10}\\MP=\sqrt{\left(-3\right)^2+\left(-1\right)^2}=\sqrt{10}\end{matrix}\right.\) 

\(\Rightarrow S_{MNP}=\dfrac{1}{2}MN.MP=5\)

ai giúp mình câu b với 

a: \(AB=\sqrt{\left[2-\left(-2\right)\right]^2+\left(-1-2\right)^2}=5\)

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

Do đó: AB=BC

hay ΔABC cân tại B

giúp mình với

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

\(BC=\sqrt{3^2+4^2}=5\left(cm\right)\)

Do đó: ΔABC cân tại B

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

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

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

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

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

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

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

=>C=3 căn 5+căn 29