1: A(1;2); B(0;-1); C(-2;3)
\(\overrightarrow{BC}=\left(-2;4\right)=\left(-1;2\right)\)
=>Vecto pháp tuyến là (2;1)
Phương trình đường thẳng BC là:
2(x-0)+1(y+1)=0
=>2x+y+1=0
Vì AH\(\perp\)BC nên AH nhận \(\overrightarrow{BC}=\left(-1;2\right)\) làm vecto pháp tuyến
Phương trình đường cao AH là:
-1(x-1)+2(y-2)=0
=>-x+1+2y-4=0
=>-x+2y-3=0
2: Tọa độ H là:
\(\left\{{}\begin{matrix}-x+2y-3=0\\2x+y+1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-x+2y=3\\2x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x+2y=3\\4x+2y=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=1\\x-2y=-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{1}{3}\\2y=x+3=\dfrac{1}{3}+3=\dfrac{10}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{5}{3}\end{matrix}\right.\)
Vậy: A(1;2); H(1/3;5/3)
\(AH=\sqrt{\left(\dfrac{1}{3}-1\right)^2+\left(\dfrac{5}{3}-2\right)^2}=\dfrac{\sqrt{5}}{3}\)
\(BC=\sqrt{\left(-2-0\right)^2+\left(3+1\right)^2}=\sqrt{2^2+4^2}=2\sqrt{5}\)
Xét ΔABC có AH là đường cao
nên \(S_{ABC}=\dfrac{1}{2}\cdot AH\cdot BC=\dfrac{1}{2}\cdot2\sqrt{5}\cdot\dfrac{\sqrt{5}}{3}=\dfrac{1}{3}\)
3: Gọi I(x;y) là tọa độ tâm đường tròn ngoại tiếp ΔABC
=>IA=IB=IC
I(x;y); A(1;2); B(0;-1); C(-2;3)
\(IA^2=\left(x-1\right)^2+\left(y-2\right)^2\)
\(IB^2=\left(x-0\right)^2+\left(y+1\right)^2=x^2+\left(y+1\right)^2\)
\(IC^2=\left(x+2\right)^2+\left(y-3\right)^2\)
Theo đề, ta có: \(\left\{{}\begin{matrix}IA^2=IB^2\\IB^2=IC^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x-1\right)^2+\left(y-2\right)^2=x^2+\left(y+1\right)^2\\x^2+\left(y+1\right)^2=\left(x+2\right)^2+\left(y-3\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-2x+1+y^2-4y+4=x^2+y^2+2y+1\\x^2+y^2+2y+1=x^2+4x+4+y^2-6y+9\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2x-4y+5=2y+1\\2y+1=4x-6y+13\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-4x-4y=-4\\4x-6y+13=2y+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y=1\\4x-8y=-12\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y=1\\x-2y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y=4\\x+y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{4}{3}\\x=1-y=1-\dfrac{4}{3}=-\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(I\left(-\dfrac{1}{3};\dfrac{4}{3}\right);A\left(1;2\right)\)
\(IA=\sqrt{\left(1+\dfrac{1}{3}\right)^2+\left(2-\dfrac{4}{3}\right)^2}=\dfrac{2\sqrt{5}}{3}\)
Phương trình đường tròn tâm I là:
\(\left(x+\dfrac{1}{3}\right)^2+\left(y-\dfrac{4}{3}\right)^2=IA^2=\dfrac{20}{9}\)
