b: Phương trình hoành độ giao điểm là:
2x-3=7-3x
=>2x+3x=7+3
=>5x=10
=>x=2
Thay x=2 vào y=2x-3, ta được:
\(y=2\cdot2-3=1\)
Vậy: A(2;1)
c: Tọa độ B là:
\(\left\{{}\begin{matrix}y=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=0\end{matrix}\right.\)
Tọa độ C là:
\(\left\{{}\begin{matrix}y=0\\-3x+7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=\dfrac{7}{3}\end{matrix}\right.\)
Ta có: A(2;1); B(3/2;0); C(7/3;0)
\(AB=\sqrt{\left(\dfrac{3}{2}-2\right)^2+\left(0-1\right)^2}=\dfrac{\sqrt{5}}{2}\)
\(AC=\sqrt{\left(\dfrac{7}{3}-2\right)^2+\left(0-1\right)^2}=\dfrac{\sqrt{10}}{3}\)
\(BC=\sqrt{\left(\dfrac{7}{3}-\dfrac{3}{2}\right)^2+\left(0-0\right)^2}=\dfrac{5}{6}\)
Xét ΔABC có \(cosBAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)
\(=\dfrac{\dfrac{5}{4}+\dfrac{10}{9}-\dfrac{25}{36}}{2\cdot\dfrac{\sqrt{5}}{2}\cdot\dfrac{\sqrt{10}}{3}}=\dfrac{\sqrt{2}}{2}\)
=>\(sinBAC=\sqrt{1-cos^2BAC}=\dfrac{\sqrt{2}}{2}\)
Diện tích tam giác ABC là:
\(S_{BAC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinBAC\)
\(=\dfrac{1}{2}\cdot\dfrac{\sqrt{5}}{2}\cdot\dfrac{\sqrt{10}}{3}\cdot\dfrac{\sqrt{2}}{2}=\dfrac{5}{12}\)