Gọi pt đường thẳng có dạng \(y=ax+b\)
a/ \(\left\{{}\begin{matrix}a=\frac{3}{2}\\-a+b=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{3}{2}\\b=\frac{9}{2}\end{matrix}\right.\) \(\Rightarrow y=\frac{3}{2}x+\frac{9}{2}\)
b/ \(3x-2y=1\Rightarrow y=\frac{3}{2}x-\frac{1}{2}\) \(\Rightarrow a=\frac{3}{2}\)
Giống như câu trên ta được pt đường thẳng \(y=\frac{3}{2}x+\frac{9}{2}\)
c/ \(3y-2x+1=0\Rightarrow y=\frac{2}{3}x-\frac{1}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{2}{3}.a=-1\\-a+b=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-\frac{3}{2}\\b=\frac{3}{2}\end{matrix}\right.\) \(\Rightarrow y=-\frac{3}{2}x+\frac{3}{2}\)
