The equation of all lines having slope 2 which are tangent to the...

The equation of the given curve is $\mathrm{y}=\frac{1}{\mathrm{x}-3}, \mathrm{x} \neq 3$ The slope of the tangent to the given curve at any point $(\mathrm{x}, \mathrm{y})$ is given by $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1}{(\mathrm{x}-3)^{2}}$ For tangent having slope $2,$ we must have $2=\frac{-1}{(x-3)^{2}} \Rightarrow 2(x-3)^{2}=-1 \Rightarrow(x-3)^{2}=-\frac{1}{2}$ which is not possible as square of a real number cannot be negative. Hence, there is no tangent to the given curve having slope 2.