Ta có \(y=-\dfrac{x}{2}+2=-\dfrac{1}{2}x+2\)
\(A\left(-1;3\right)\inđths;đths\text{//}y=-\dfrac{1}{2}x+2\Leftrightarrow\left\{{}\begin{matrix}-a+b=3\\a=-\dfrac{1}{2};b\ne2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=\dfrac{5}{2}\end{matrix}\right.\)