Question 166145
since (1,2) and (4,-1) are the vertices of the isosceles and the other vertex is (6,y) we know that the distance between (1,2) and the 3rd vertex will be equal to the distance between (4,-1) and the 3rd vertex.

using the distance formula {{{sqrt((6-1)^2+(y-2)^2)=sqrt((6-4)^2+(y+1))^2}}}--->
square both sides to get rid of squareroot signs then simplify

25+y^2-4y+4=4+y^2+2y+1----->6y=24---{{{y=4}}} so the 3rd vertex is (6,4)