a:
Bảng giá trị:
x | 0 | 1 |
y=2x+1 | 1 | 3 |
y=-2x+4 | 4 | 2 |
Vẽ đồ thị:
b; Tọa độ C là:
\(\begin{cases}2x+1=-2x+4\\ y=2x+1\end{cases}\Rightarrow\begin{cases}4x=3\\ y=2x+1\end{cases}\Rightarrow\begin{cases}x=\frac34\\ y=2\cdot\frac34+1=\frac32+1=\frac52\end{cases}\)
=>C(3/4;5/2)
Tọa độ A là:
\(\begin{cases}y=0\\ 2x+1=0\end{cases}\Rightarrow\begin{cases}y=0\\ 2x=-1\end{cases}\Rightarrow\begin{cases}y=0\\ x=-\frac12\end{cases}\)
=>A(-1/2;0)
Tọa độ B là:
\(\begin{cases}y=0\\ -2x+4=0\end{cases}\Rightarrow\begin{cases}y=0\\ -2x=-4\end{cases}\Rightarrow\begin{cases}y=0\\ x=2\end{cases}\)
=>B(2;0)
c: C(3/4;5/2); A(-1/2;0); B(2;0)
\(CA=\sqrt{\left(-\frac12-\frac34\right)^2+\left(0-\frac52\right)^2}=\sqrt{\left(-\frac54\right)^2+\left(-\frac52\right)^2}\)
\(=\sqrt{\frac{25}{16}+\frac{25}{4}}=\sqrt{\frac{25}{16}+\frac{100}{16}}=\sqrt{\frac{125}{16}}=\frac{5\sqrt5}{4}\)
\(CB=\sqrt{\left(2-\frac34\right)^2+\left(0-\frac52\right)^2}=\sqrt{\left(\frac54\right)^2+\left(\frac52\right)^2}=\frac{5\sqrt5}{4}\)
\(AB=\sqrt{\left(2+\frac12\right)^2+\left(0-0\right)^2}=\frac52\)
Chu vi tam giác ABC là:
\(C_{ABC}=AB+AC+BC\)
\(=\frac{5\sqrt5}{4}+\frac{5\sqrt5}{4}+\frac52=\frac{10\sqrt5}{4}+\frac52=\frac{5\sqrt5+5}{2}\)
Xét ΔABC có
\(cosC=\frac{CA^2+CB^2-AB^2}{2\cdot CA\cdot CB}\)
\(=\left(\frac{125}{16}+\frac{125}{16}-\frac{25}{4}\right):\left(2\cdot\frac{5\sqrt5}{4}\cdot\frac{5\sqrt5}{4}\right)=\left(\frac{125}{8}-\frac{25}{4}\right):\frac{2\cdot25\cdot5}{16}\)
\(=\frac{75}{8}\cdot\frac{16}{50\cdot5}=\frac{75}{50\cdot5}\cdot2=\frac{3}{2\cdot5}\cdot2=\frac35\)
=>\(\sin C=\sqrt{1-\left(\frac35\right)^2}=\frac45\)
Diện tích tam giác CAB là:
\(S_{CAB}=\frac12\cdot CA\cdot CB\cdot\sin C\)
\(=\frac12\cdot\frac{5\sqrt5}{4}\cdot\frac{5\sqrt5}{4}\cdot\frac45=\frac12\cdot\frac{25\cdot5}{16}\cdot\frac45=\frac{1}{32}\cdot25\cdot4=\frac{100}{32}=3,125\)
