Upon reflection

Let point $P'=(x',y')$ be the reflection of point $P=(x,y)$ on $x+2y-4=0$ in the line $x-y-2=0$ .

We notice that the midpoint of $PP'$ lies on $x-y-2=0$ , so we have $\frac{x+x'}{2}-\frac{y+y'}{2}-2=0 \Rightarrow x-y=-x'+y'+4$ .

$PP'$ is also perpendicular to $x-y-2=0$ . Since the slope of $x-y-2=0$ is $1$ , $PP'$ must have slope $-1$ . Thus, $\frac{y-y'}{x-x'}=-1 \Rightarrow x+y=x'+y'$ .

Solving the above two equations for $x$ and $y$ , we have $x=y'+2$ and $y=x'-2$ . Since point $(x,y)$ lies on $x+2y-4=0$ , substituting these into $x+2y-4=0$ , we have $y'+2+2(x'-2)-4=0 \Rightarrow y'=-2x'+6.$

Therefore, $a=-2$ and $b=6$ , hence $a+b = 4$ .