The correct options are
B (5,5)
C (2,2)
Distance of point from (3,4) is √5 and abcissa and ordinate is same then let say x=y=a.
Now using distance formula, we have
√(a−3)2+(a−4)2=√5
(a−3)2+(a−4)2=5
a2+9−6a+a2+16−8a=5
2a2−14a+20=0
a2−7a+10=0
(a−5)(a−2)=0
a=5 and 2
Points are (2,2) and (5,5).