If the point $A (2, -4)$ is equidistant from $P (3, 8)$ and $Q (-10, y)$,find the values of $y$. Also,find the distance $PQ$.

(N/A) According to the question,point $A (2, -4)$ is equidistant from $P (3, 8)$ and $Q (-10, y)$.
This means $PA = QA$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$,we have:
$\sqrt{(2 - 3)^2 + (-4 - 8)^2} = \sqrt{(2 - (-10))^2 + (-4 - y)^2}$
$\sqrt{(-1)^2 + (-12)^2} = \sqrt{(12)^2 + (-(4 + y))^2}$
$\sqrt{1 + 144} = \sqrt{144 + (4 + y)^2}$
$\sqrt{145} = \sqrt{144 + 16 + 8y + y^2}$
Squaring both sides:
$145 = 160 + 8y + y^2$
$y^2 + 8y + 15 = 0$
$(y + 5)(y + 3) = 0$
So,$y = -5$ or $y = -3$.
Now,calculating distance $PQ = \sqrt{(-10 - 3)^2 + (y - 8)^2} = \sqrt{(-13)^2 + (y - 8)^2} = \sqrt{169 + (y - 8)^2}$.
For $y = -3$,$PQ = \sqrt{169 + (-3 - 8)^2} = \sqrt{169 + 121} = \sqrt{290}$.
For $y = -5$,$PQ = \sqrt{169 + (-5 - 8)^2} = \sqrt{169 + 169} = \sqrt{338} = 13\sqrt{2}$.