Question

Trong mpOxy cho tam giác ABC với A(1 ; 5) B(3;–1) C(6;0). Tìm chân đường cao B’ kẻ từ B lên CA.

Answer 1

Gọi \(B'\left(x;y\right)\)

Ta có: \(\overrightarrow{BB'}=\left(x-3;y+1\right)\) ; \(\overrightarrow{AC}=\left(5;-5\right)\) ; \(\overrightarrow{AB'}=\left(x-1;y-5\right)\)

Do \(\left\{{}\begin{matrix}BB'\perp AC\\B'\in AC\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5\left(x-1\right)-5\left(y+1\right)=0\\\frac{x-1}{5}=\frac{y-5}{-5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=2\\x+y=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\)

\(\Rightarrow B'\left(4;2\right)\)