A (4,3)
B (-5,6)
C (-4,-1)
Gọi H(x;y)
\(\overrightarrow{AH}=\left(x-4;y-3\right)\)
\(\overrightarrow{BC}=\left(1;-7\right)\)
\(\overrightarrow{BH}=\left(x+5;y-6\right)\)
H là trực tâm của tam giác ABC
\(\Leftrightarrow\left\{{}\begin{matrix}\overrightarrow{AH.}\overrightarrow{BC}=\overrightarrow{0}\\\overrightarrow{BH},\overrightarrow{BC}\end{matrix}\right.\)
cùng phương
\(\Leftrightarrow\left\{{}\begin{matrix}1.\left(x-4\right)+\left(-7\right)\left(y-3\right)=0\\\dfrac{x+5}{1}=\dfrac{y-6}{-7}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-4-7y+21=0\\-7\left(x+5\right)=y-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-7y=-17\\-7x-35=y-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-7y=-17\\-7x-y=29\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-22}{5}\\y=\dfrac{9}{5}\end{matrix}\right.\)
Vậy \(H\left(\dfrac{-22}{5};\dfrac{9}{5}\right)\)
mình làm lộn với chân đường cao r, xin lỗi bạn nha, mình sửa lại nè
Gọi H (x;y)
\(\overrightarrow{AH}=\left(x-4;y-3\right)\)
\(\overrightarrow{BC}=\left(1;-7\right)\)
\(\overrightarrow{BH}=\left(x+5;y-6\right)\)
\(\overrightarrow{AC}=\left(-8;-4\right)\)
H là trực tâm của tam giác ABC
\(\Leftrightarrow\left\{{}\begin{matrix}\overrightarrow{AH}.\overrightarrow{BC}=\overrightarrow{0}\\\overrightarrow{BH}.\overrightarrow{AC}=\overrightarrow{0}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-4\right)-7\left(y-4\right)=0\\-8\left(x+5\right)-4\left(y-6\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-4-7y+28=0\\-8x-40-4y+24=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-7y=-24\\-8x-4y=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-52}{15}\\y=\dfrac{44}{15}\end{matrix}\right.\)
Vậy \(H\left(\dfrac{-52}{15};\dfrac{44}{15}\right)\)
