Tọa độ A là:
\(\left\{{}\begin{matrix}y=0\\\dfrac{3}{2}x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=2\end{matrix}\right.\)
Tọa độ B là:
\(\left\{{}\begin{matrix}y=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=3\end{matrix}\right.\)
Tọa độ C là:
\(\left\{{}\begin{matrix}\dfrac{3}{2}x-3=x-3\\y=x-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{2}x=0\\y=x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0-3=-3\end{matrix}\right.\)
Vậy: A(2;0); B(3;0); C(0;-3)
\(AB=\sqrt{\left(3-2\right)^2+\left(0-0\right)^2}=1\)
\(AC=\sqrt{\left(0-2\right)^2+\left(-3-0\right)^2}=\sqrt{13}\)
\(BC=\sqrt{\left(0-3\right)^2+\left(-3-0\right)^2}=3\sqrt{2}\)
Xét ΔABC có \(cosBAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)
\(=\dfrac{1+13-18}{2\cdot1\cdot\sqrt{13}}=\dfrac{-4}{2\sqrt{13}}=-\dfrac{2}{\sqrt{13}}\)
=>\(sinBAC=\sqrt{1-\left(-\dfrac{2}{\sqrt{13}}\right)^2}=\dfrac{3}{\sqrt{13}}\)
Diện tích tam giác ABC là:
\(S_{BAC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinBAC\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{\sqrt{13}}\cdot1\cdot\sqrt{13}=\dfrac{3}{2}\)
