Question 892474
You want your unknown point (6,y) to be the same distance from (1,2) as it is from (4,-1).
The segments, (6,y) to (1,2), and (6,y) to (4,-1), are the two equal sides of the isosceles
triangle.


{{{sqrt((6-1)^2+(y-2)^2)=sqrt((6-4)^2+(y-(-1))^2)}}}
{{{sqrt((5)^2+(y-2)^2)=sqrt((2)^2+(y+1)^2)}}}
{{{sqrt(25+(y-2)^2)=sqrt(4+(y+1)^2)}}}
{{{(y-2)^2+25=(y+1)^2+4}}}
{{{y^2-4y+4+25=y^2+2y+1+4}}},  see {{{y^2}}} on both sides
{{{-4y+29=2y+5}}}
{{{-6y+29=5}}}
{{{-6y=5-29}}}
{{{6y=29-5}}}
{{{6y=24}}}
{{{highlight(y=4)}}} --------the ordinate, or y value for the vertex opposite of the base.