L1:4x+3y−12=0
L2:3x+4y−12=0
L1+λL2=0
⇒(4x+3y−12)+λ(3x+4y−12)=0
⇒x(4+3λ)+y(3+4λ)−12(1+λ)=0
Point A(4+3λ12(1+λ),0)
Point B(0,3+4λ12(1+λ))
Mid point ⇒h=4+3λ6(1+λ) .......(i)
k=3+4λ6(1+λ) .......(ii)
Eliminate λ from equations (i) and (ii) then
6(h+k)=7hk
Hence, locus is
6(x+y)=7xy