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

Step-by-step explanation:

10^2={(10-2)^2+(y+3)^2

100-64= 9+y^2+6y

36-9=y^2+6y

y^2+6y-27=0

y^2+9y-3y-27=0

y(y+9) -3(y+9) =0

(y-3) (y+9) =0

y=3, -9