The given function is expressed as, |PA−PB|=3 √(h−1)2+(k−2)2−√(h−2)2+(k−1)2=3 √(h−1)2+(k−2)2=3+√(h−2)2+(k−1)2 Square both sides of the above expression and simplify it, 2h−2k−9=6√(h−2)2+(k−1)2 Again square the above expression, then (2h−2k−9)2=36[(h−2)2+(k−1)2] 32h2+32k2+8hk−108h−108k+99=0 Thus, the locus of p(h,k) is, 32x2+32y2+8xy−108x−108y+99=0