Step 1: Idea of a mirror image.
The image $(h,k)$ of the point $(1,2)$ in the mirror line is such that the line joining the point to its image is perpendicular to the mirror, and the midpoint of that join lies on the mirror. We use these two facts directly.
Step 2: Slope condition.
The line $x-3y+4=0$ has slope $\tfrac{1}{3}$. The segment from $(1,2)$ to $(h,k)$ must be perpendicular to it, so its slope is $-3$. That gives $\dfrac{k-2}{h-1}=-3$, i.e. $k-2=-3(h-1)$, so $3h+k=5$.
Step 3: Midpoint on the line.
The midpoint is $\left(\dfrac{h+1}{2},\dfrac{k+2}{2}\right)$. It lies on $x-3y+4=0$, so $\dfrac{h+1}{2}-3\cdot\dfrac{k+2}{2}+4=0$.
Step 4: Clear the fractions.
Multiply by $2$: $(h+1)-3(k+2)+8=0$, which gives $h-3k+3=0$, i.e. $h-3k=-3$.
Step 5: Solve the two equations.
We have $3h+k=5$ and $h-3k=-3$. From the first, $k=5-3h$. Put into the second: $h-3(5-3h)=-3$, so $h-15+9h=-3$, giving $10h=12$ and $h=\tfrac{6}{5}$.
Step 6: Find $k$ and state the image.
$k=5-3\cdot\tfrac{6}{5}=5-\tfrac{18}{5}=\tfrac{7}{5}$. So the image is $\left(\tfrac{6}{5},\tfrac{7}{5}\right)$. \[ \boxed{\left(\tfrac{6}{5},\ \tfrac{7}{5}\right)} \]