It is given that, AP+BP=2 Consider the point P be (x,y) √(x−1)2+y2+√x2+(y−1)2=2 √(x−1)2+y2=2−√x2+(y−1)2 Squaring on both sides, (x−1)2+y2=4+x2+(y−1)2−4√x2+(y−1)2 x2+y2+1−2x=4+x2+y2+1−2y−4√x2+(y−1)2 2y−2x=4−4√x2+(y−1)2 y−x−2=−2√x2+(y−1)2 Squaring both the sides x2+y2+4−2xy+4x−4y=4x2+4y2+4−8y 3x2+3y2−4x−4y+2xy=0