If the equation of one asymptote of the hyperbola $14{\mathrm{x}}^2 + 38\mathrm{xy} + 20{\mathrm{y}}^2 + \mathrm{x} – 7\mathrm{y} – 91=0$ is $7\mathrm{x} + 5\mathrm{y} – 3 = 0$ then the other asymptote =

$$\begin{gathered} \mathrm{S}\equiv 14{\mathrm{x}}^2+38\mathrm{xy}+20{\mathrm{y}}^{2\backslash }+\mathrm{x}-7\mathrm{y}-91=0 \\ Now \frac{\partial S}{\partial x}=0\Rightarrow 28x+38y+1=0\to \left(1\right) and \\ \frac{\partial S}{\partial y}=0\Rightarrow 38x+40y-7=0\to \left(2\right) \\ point of intersection of \left(1\right) and \left(2\right) is centre \\ \therefore \mathrm{centre} \mathrm{of} \mathrm{hyperbola}=\left(\frac{17}{18}, \frac{-13}{18}\right) \\ \mathrm{we} \mathrm{know} \mathrm{that} \mathrm{asymptotes} \mathrm{are} \mathrm{passes} \mathrm{through} \mathrm{centre} \mathrm{of} \mathrm{hyperbola} \\ \therefore \mathrm{By} \mathrm{verification} 2^{\mathrm{nd}} \mathrm{option} 2\mathrm{x}+4\mathrm{y}+1=0 satisfies the point \left(\frac{17}{18}, \frac{-13}{18}\right) \end{gathered}$$