\(\overrightarrow{CA}=\left(3;3\right)=3\left(1;1\right)\Rightarrow\) đường thẳng AC nhận \(\left(1;-1\right)\) là 1 vtpt
Phương trình AC:
\(1\left(x-1\right)-1\left(y-4\right)=0\Leftrightarrow x-y+3=0\)
b/ \(\overrightarrow{AB}=\left(2;-4\right)=2\left(1;-2\right)\)
\(CK\perp AB\Rightarrow CK\) nhận \(\left(1;-2\right)\) là 1 vtpt
Phương trình CK:
\(1\left(x+2\right)-2\left(y-1\right)=0\Leftrightarrow x-2y+4=0\)
c/ M là trung điểm AC \(\Rightarrow M\left(-\frac{1}{2};\frac{5}{2}\right)\Rightarrow\overrightarrow{BM}=\left(-\frac{5}{2};\frac{5}{2}\right)=-\frac{5}{2}\left(1;-1\right)\)
Đường thẳng BM nhận (1;1) là 1 vtpt
Pt BM: \(1\left(x-2\right)+1\left(y-0\right)=0\Leftrightarrow x+y-2=0\)
d/ \(AC=\sqrt{3^2+3^2}=3\sqrt{2}\) ; \(AB=\sqrt{2^2+\left(-4\right)^2}=2\sqrt{5}\)
\(\overrightarrow{BC}=\left(-4;1\right)\Rightarrow BC=\sqrt{4^2+1^2}=\sqrt{17}\)
\(cosC=\frac{BC^2+AC^2-AB^2}{2BC.AC}=\frac{5\sqrt{34}}{68}\)
\(\Rightarrow C\approx64^036'\)
