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

The distance between P(2,-3) and Q(10,y) is 10 units.
⇒ √((10-2)²+[y-(-3)]²) = 10
⇒ √(64+y²+9+6y) = 10
Squaring both sides
64+y²+9+6y = 100
⇒ y²+6y-27 = 0
⇒ y²+9y-3y-27 = 0
⇒ y(y+9)-3(y+9) = 0
⇒ (y+9)(y-3) = 0
⇒ (y+9) = 0 or (y-3) = 0
⇒ y = -9 or y = 3