one of the bisector of the angle between the lines a(x-1)²+2h(x-1)(y-2)+b(y-2)² =0 is

x+2y-5 = 0 the other bisector is ??answer is 2x-y =0

explain ??

since the two bisectors are perpendicular.
the equation of the given bisector is $x+2y-5=0$
$i.e. y=-\frac{1}{2}x+\frac{5}{2}$
the slope of the given bisector is $-\frac{1}{2}$ .

therefore the slope of the other bisector is 2.
the given equation of pair of straight lines is $a(x-1{)}^2+2h(x-1\left)\right(y-2)+b(y-2{)}^2=0 ...........\left(1\right)$
the intersection point of the pair of straight line is given by substituting $x-1=0 and y-2=0$
$i.e. x=1 and y=2$ .
the lines are concurrent at (1,2).
the angle bisector of the straight lines will also pass through the point (1,2).

therefore the other bisector passes through the point (1,2) and slope is 2.
thus the equation of the other bisector is
$$\begin{gathered} y-2=2*(x-1) \\ y-2=2x-2 \\ \Rightarrow 2x-y=0 \end{gathered}$$

hope this helps you