Question

find the values of y for which the distance between the points P(2,-3) and Q(10,y) is 10 units

Answer 1

square root of (10-2)^2+(-3-y)^2=10
squaring on both sides
100=(8)^2+(3+y)^2
100-64=9+6y+y^2
y^2+6y-27
y^2+9y-3y-27
y(y+9)-3(y+9)
(y-3)(y+9)
then y=3,-9
substituting y=-9 we do not get answer 
then y=3