If slope of the tangent to the curve \(x y+a x+b y=0\) at the point \((1,1)\) on it is 2 , then the value of \(3 a+b\) is

\(x y+a x+b y=0\)
Differentiating w.r.t. \(x\), we get
\(\frac{x \mathrm{~d} y}{\mathrm{~d} x}+y+\mathrm{a}+\mathrm{b} \frac{\mathrm{d} y}{\mathrm{~d} x}=0\)
\(\begin{aligned} & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{\mathrm{a}+y}{\mathrm{~b}+x} \\ & \Rightarrow\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{(\mathrm{b}, 1)}=-\frac{\mathrm{a}+1}{\mathrm{~b}+1} \\ & \Rightarrow 2=-\frac{\mathrm{a}+1}{\mathrm{~b}+1}\end{aligned}\)
\(\Rightarrow a+2 b=-3\) ...(i)
Also, the point \((1,1)\) lies on the curve
\(x y+a x+b y=0\).
\(\therefore \quad 1+a+b=0\)
\(\Rightarrow a+b=-1\) ...(ii)
Solving (i) and (ii), we get \(a=1, b=-2\)
\(\therefore \quad 3 a+b=1\)