\(cosB=\dfrac{\left|1.2+\left(-7\right).1\right|}{\sqrt{1^2+\left(-7\right)^2}.\sqrt{2^2+1^2}}=\dfrac{1}{\sqrt{10}}\)
Gọi vtpt của AC có tọa độ \(\left(a;b\right)\)
\(\Rightarrow cosC=cosB=\dfrac{1}{\sqrt{10}}=\dfrac{\left|2a+b\right|}{\sqrt{a^2+b^2}.\sqrt{2^2+1^2}}=\dfrac{1}{\sqrt{10}}\)
\(\Leftrightarrow\sqrt{2}\left|2a+b\right|=\sqrt{a^2+b^2}\)
\(\Leftrightarrow2\left(2a+b\right)^2=a^2+b^2\)
\(\Leftrightarrow7a^2+8ab+b^2=0\Leftrightarrow\left(a+b\right)\left(7a+b\right)=0\)
Chọn \(a=1\Rightarrow\left[{}\begin{matrix}b=-1\\b=-7\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left(a;b\right)=\left(1;-1\right)\\\left(a;b\right)=\left(1;-7\right)\end{matrix}\right.\)
(Trường hợp \(\left(a;b\right)=\left(1-;7\right)\) loại do khi đó AC song song AB, vô lý)
\(\Rightarrow\) Phương trình AC: \(1\left(x-4\right)-1\left(y-0\right)=0\)