If  $2x+y-4=0$ is a bisector of angles between the lines $a\left(x-1\right)+b\left(y-2\right)=0,$ $c\left(x-1\right)+d\left(y-2\right)=0$ then other angular bisector is 

$a\left(x-1\right)+b\left(y-2\right)=0,$ $c\left(x-1\right)+d\left(y-2\right)=0$ 

P(1, 2) lie on both the lines and hence P(1, 2)  is their point of intersection.

$\text{ One of the bisector is }2x+y-4=0$
Other bisector will be perpendicular to this bisector.

Hence its equation will be$x-2y+k=0\to \left(1\right)$

Passes through $\left(1,2\right)$

$1-4+k=0\Rightarrow k=3$
$x-2y+3=0$