The coordinates of the point P on the line 2x+3y+1=0 , such that |PA−PB| is maximum, where A is (2, 0) and B is (0, 2) is

|PA−PB|≤|AB| Maximum value of |PA−PB|  is  |AB| , which is possible only when P, A, B are collinear.If P (x, y) then equation AB is x2+y2=1⇒ x+y=2    ….(i)            Now solving Eq. (i) and                 2x+3y+1=0   ….(ii)There we get,  x=7,  y=−5                  ∴        P=(7,−5)