Suppose A,B are two points on 2x−y+3=0 and P(1,2) is such that PA=PB , then the  mid-point of AB is

Equation of AB is 2x−y+3=0 We have ΔPAD≅△PBD ⇒D is foot of perpendicular from P to AB ⇒α−12=β−2−1=−[2×1−1×2+3]4+1⇒α−12=β−2−1=−35α=−15;β=135D=(α,β)=−15,135