One of the bisectors of the angle between the lines  $a{\left(x-1\right)}^2+2b\left(x-1\right)\left(y-2\right)+b{\left(y-2\right)}^2=0$is $x+2y-5=0$ then the value of $\frac{b-a}{h}$ is______.

Shifting the origin to (1,2), the equation of given pair of lines becomes

$a{x}^2+2hxy+b{y}^2=0$

Whose one angle bisector is $x+2y=0$(given) and $2x-y=0$

$\therefore$combined equation of angle bisector is $\left(x+2y\right)\left(2x-y\right)=0$
$2x^2+3xy-2y^2=0$

On comparing with $\frac{x^2-y^2}{a-b}=\frac{xy}{h}$, we get
$$\begin{gathered} \frac{x^2-y^2}{4}=\frac{2xy}{3}\Rightarrow 3x^2-8xy-3y^2=0 \\ a=3, b=-3, h=-4 \end{gathered}$$

$a-b=-3$and h=2,

$\frac{b-a}{h}=\frac{3}{2}=1.5$