The equation of a straight line passing through the points (1, 2) and inclined at ${45}^0$to the line $y=2x+1$ is

Let m be the slope of the line passing through the point (1, 2)

Slope of $y=2x+1$ is '2'

Let $m_1=m, m_2=2 and \theta ={45}^0$

     $$\begin{gathered} Tan {45}^0=\left|\frac{m-2}{1+2m}\right| \\ \Rightarrow \left|\frac{m-2}{1+2m}\right|=1 \\ \Rightarrow \frac{m-2}{1+2m} =\pm 1 \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{m-2}{1+2m}=1 or \frac{m-2}{1+2m}=-1 \\ \Rightarrow m-2=1+2m or m-2=-1-2m \\ \Rightarrow m=-3 or m=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right. \end{gathered}$$

If $m=-3$ then equation is 

$$\begin{gathered} y-2=-3\left(x-1\right) \\ \Rightarrow 3x+y-5=0 \end{gathered}$$

If $m=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ then equation is

$$\begin{gathered} y-2=\frac{1}{3}\left(x-1\right) \\ \Rightarrow 3y-6=x-1 \\ \Rightarrow x-3y+5=0 \end{gathered}$$