The correct option is C (2±√112,52)
Given−A(2,0),B(2,5)&C(x,y)aretheverticesofanisoscelesΔABC.Tofindout−C(x,y)=?Solution−Usingdistanceformula,AB=√(2−2)2+(0−5)2units=5units≠3units.∴ABisthebaseofthegiventriangle.∴AC=BC=3units.........(i)⟹√(x−2)2+(y−0)2=√(x−2)2+(y−5)2⟹−10y+25=0⟹y=52.puttingy=52inAgainAC=BC=3units∴√(x−2)2+(y−0)2=3(byi)⟹x2+y2−4x+4=9⟹.x2+y2−4x−5=0Puttingy=52,x2+(52)2−4x−5=0⟹4x2−16x+5=0⟹x=16±√162−4×4×52⟹x=2±√112.∴C(x,y)=(2±√112,52).Note−the±signofxand+signofyshowsthatthetriangleisinthefirstaswellasinthesecondquadrant.Ans−OptionC.