\(C\in CK\Rightarrow C\left(x;-\dfrac{3}{8}x-\dfrac{13}{8}\right)\)
\(\Rightarrow\overrightarrow{BC}=\left(x+4;-\dfrac{3}{8}x-\dfrac{53}{8}\right)\)
AH có VTPT là \(\overrightarrow{n}=\left(5;3\right)\)
Do \(AH\) vuông góc \(BC\Rightarrow\overrightarrow{BC}=k\overrightarrow{n}\)
\(\Rightarrow\left\{{}\begin{matrix}x+4=5k\\-\dfrac{3}{8}x-\dfrac{53}{8}=3k\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{361}{39}\\k=-\dfrac{41}{39}\end{matrix}\right.\Rightarrow C\left(-\dfrac{361}{39};\dfrac{24}{13}\right)\).
\(A\in AH\Rightarrow A\left(x;-\dfrac{5}{3}x+\dfrac{4}{3}\right)\)
\(\Rightarrow\overrightarrow{BA}=\left(x+4;-\dfrac{5}{3}x-\dfrac{11}{3}\right)\)
\(CK\) có VTPT \(\overrightarrow{n}=\left(3;8\right)\)
Do \(CK\) vuông góc \(AB\Rightarrow\overrightarrow{BA}=k\overrightarrow{n}\)
\(\Rightarrow\left\{{}\begin{matrix}x+4=3k\\-\dfrac{5}{3}x-\dfrac{11}{3}=8k\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{43}{13}\\k=\dfrac{3}{13}\end{matrix}\right.\Rightarrow A\left(-\dfrac{43}{13};\dfrac{89}{13}\right)\).
