Question

When the distance between the points P(2,-3) and Q(10,y) is 10 units, the value of ‘y’ is__​

Answer 1

Answer:

I am not understanding your question is

Answer 2

Answer:

PQ = √(2-10)^2 + (-3-y)^2

10 = √(-8)^2 + (-3)^2 + y^2 - 2(-3)(y)

10^2 = 64 + 9 + y^2 + 6y

100 = 73 + y^2 + 6y

100-73=y^2 + 6y

27 = y^2 + 6y

y^2 + 6y - 27 = 0

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

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

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

y=-9 or y=3