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

Using Distance formula, we have
\(\Rightarrow 10 = \sqrt{(2 -10)^2 + (-3-y)^2}\)
\(\Rightarrow 10 = \sqrt{(-8)^2 + 9 + y^2 + 6y}\)
\(\Rightarrow 10 = \sqrt{64 + 9 + y^2 + 6y}\)
Squaring both sides, we get
\(\Rightarrow 100 = 73 + y^2 + 6y\)
\(\Rightarrow y^2 + 6y + 27 = 0\)

Solving this Quadratic equation by factorization, we can write
\(\Rightarrow y^2 + 9y - 3y - 27 = 0\)
\( \Rightarrow y (y + 9) – 3 (y + 9) = 0\)
\(\Rightarrow (y + 9) (y - 3) = 0\)
\(\Rightarrow y = 3, -9\)