The point of intersection of the pair of lines \(x^2+x y+2 y^2-3 x+2 y+4=0\) is

The point of intersection of the pair of lines \(f(x, y) \equiv x^2+x y+2 y^2-3 x+2 y+4=0\) is same as the intersection of curve obtaining after partial differentiating of the curve w.r.t. \(x\) and \(y\) respectively. \(\begin{aligned} \because & \frac{\partial f}{\partial x} =2 x+y-3=0 \quad \ldots (i) \\ \text {and } & \frac{\partial f}{\partial y} =x+4 y+2=0 \quad \ldots (ii) \end{aligned}\) By cross-multiplication method, we get \(\begin{array}{rlrl} & & \frac{x}{2+12} & =\frac{-y}{4+3}=\frac{1}{8-1} \\ \Rightarrow & (x, y) & =(2,-1) \end{array}\) Hence, option (d) is correct.