Question 1124692
Let  the intersection at X axis be (0,y)

PA =PB  Any point on the perpendicular bisector is equidistant from the endpoints

Distance formula

{{{d= sqrt((x1-x2)^2+(y1-y2)^2)}}}


PA^2= PB^2

((1-0)^2+(2-y)^2)= (3-0)^2+(4-y)^2

1+4-4y+y^2=9+16-8y+y^2

rearrange

8y-4y = 25-5

4y = 20
y= 5

PA^2 =((1-0)^2+(2-5)^2)=10
PA= sqrt(10)

*[illustration bisector.png].