Tọa độ A là:
\(\left\{{}\begin{matrix}2x-3y-1=0\\5x-2y+1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-3y=1\\5x-2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10x-15y=5\\10x-4y=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-11y=7\\5x-2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{7}{11}\\5x=2y-1=\dfrac{-14-11}{11}=-\dfrac{25}{11}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{5}{11}\\y=-\dfrac{7}{11}\end{matrix}\right.\)
Tọa độ B là:
\(\left\{{}\begin{matrix}2x-3y-1=0\\2x+3y+7=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-3y=1\\2x+3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x=-6\\2x-3y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\3y=2x-1=-3-1=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
Tọa độ C là:
\(\left\{{}\begin{matrix}2x+3y+7=0\\5x-2y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=-7\\5x-2y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+6y=-14\\15x-6y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19x=-17\\5x-2y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{17}{19}\\2y=5x+1=\dfrac{-85+19}{19}=\dfrac{-64}{19}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{17}{19}\\y=-\dfrac{32}{19}\end{matrix}\right.\)
Vậy: A(-5/11;-7/11); B(-3/2;-4/3); C(-17/19;-32/19)
AC: 5x-2y+1=0
=>VTPT là (5;-2)
=>VTCP là \(\overrightarrow{AC}=\left(2;5\right)\)
mà AC\(\perp\)BH
nên BH nhận \(\overrightarrow{b}=\left(2;5\right)\) làm vecto pháp tuyến
Phương trình đường cao BH là:
\(2\left(x+\dfrac{3}{2}\right)+5\left(y+\dfrac{4}{3}\right)=0\)
=>\(2x+3+5y+\dfrac{20}{3}=0\)
=>\(2x+5y+\dfrac{29}{3}=0\)
