Question 64975
FIND THE POINT ON THE POSITIVE y-AXIS THAT IS A DISTANCE 5 FROM THE POINT P(3,4).
{{{d=sqrt((x2-x1)^2+(y2-y1)^2)}}}
d=5, (x1,y1)=(3,4) and (x2,y2)=(0,y)
{{{5=sqrt((0-3)^2+(y-4)^2)}}}
{{{5=sqrt(9+(y-4)^2)}}}
{{{(5)^2=(sqrt(9+(y-4)^2))^2}}}
{{{25=9+(y-4)^2}}}
{{{25-9=9-9+(y-4)^2}}}
{{{16=(y-4)^2}}}
{{{+-sqrt(16)=sqrt(y-4)^2}}}
+\-4=y-4
4+\-4=y-4+4
4+\-4=y
4+4=y or 4-4=y
8=y or 0=y
Since 0 is neither positive nor negative, don't use it.  
The point is (0,8)
Happy Calculating!!!