Let the point be P(x,y) which lies on line 4x−y−2=0 . . . (i) and point P is equidistant from A(−5,6) and B(3,2). Then, PA=PB or ‌‌PA2=PB2 ⇒(x+5)2+(y−6)2=(x−3)2+(y−2)2 ⇒x2+10x+25+y2−12y−36 ‌‌=‌‌x2−6x+9+y2−4y+4 ⇒‌16x−8y+48‌‌=0 ⇒‌2x−y+6‌‌=0 . . . (ii) Solving Eqs. (i) and (ii), we get x=4,y=14