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# 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
$$\Rightarrow y^2 + 9y - 3y - 27 = 0$$
$$\Rightarrow y (y + 9) – 3 (y + 9) = 0$$
$$\Rightarrow (y + 9) (y - 3) = 0$$
$$\Rightarrow y = 3, -9$$