a) \(\overrightarrow{AB}=\left(-3;-1\right)\)
\(\overrightarrow{AC}=\left(-2;4\right)\)
\(\overrightarrow{BC}=\left(1;5\right)\)
Ta thấy \(\dfrac{-3}{-2}\ne\dfrac{-1}{4}\)
\(\Rightarrow\overrightarrow{AB};\overrightarrow{AC}\) không cùng phương
\(\Rightarrow A;B;C\) không thẳng hàng
\(\Rightarrow ABC\) là tam giác
b) Gọi \(N\left(x;y\right)\in\left(AC\right)\)
\(\overrightarrow{n_p\left(AC\right)}=\left(-4;-2\right)=\left(2;1\right)\)
\(\left(AC\right):2\left(x-0\right)+\left(y-8\right)=0\Leftrightarrow2x+y-8=0\left(1\right)\)
\(\overrightarrow{n_p\left(BC\right)}=\left(-5;1\right)\Rightarrow\left(BC\right):-5\left(x-0\right)+\left(y-8\right)=0\Leftrightarrow5x-y+8=0\)
\(\overrightarrow{n_p\left(AB\right)}=\left(1;-3\right)\Rightarrow\left(AB\right):\left(x+1\right)-3\left(y-3\right)=0\Leftrightarrow x-3y+10=0\)
\(S_{ABN}=\dfrac{1}{2}.d\left(N;AB\right).AB=\dfrac{1}{2}.\dfrac{\left|x-3y+10\right|}{\sqrt{10}}.\sqrt{10}=\dfrac{\left|x-3y+10\right|}{2}\)
\(S_{BNC}=\dfrac{1}{2}d\left(N;BC\right).BC=\dfrac{1}{2}.\dfrac{\left|5x-y+8\right|}{\sqrt{26}}.\sqrt{26}=\dfrac{\left|5x-y+8\right|}{2}\)
\(S_{ABN}=4.S_{BNC}\)
\(\Leftrightarrow\dfrac{\left|x-3y+10\right|}{2}=4.\dfrac{\left|5x-y+8\right|}{2}\)
\(\Leftrightarrow\left|x-3y+10\right|=4.\left|5x-y+8\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3y+10=4\left(5x-y+8\right)\\x-3y+10=-4\left(5x-y+8\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}19x-y+22=0\left(2\right)\\21x-7y+42=0\left(3\right)\end{matrix}\right.\)
\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=\dfrac{28}{3}\end{matrix}\right.\)
\(\left(1\right);\left(3\right)\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=\dfrac{36}{5}\end{matrix}\right.\)
Vậy \(N\left(-\dfrac{2}{3};\dfrac{28}{3}\right)\) hoặc \(N\left(\dfrac{2}{5};\dfrac{36}{5}\right)\)
