If $Q(0, 1)$ is equidistant from $P(5, -3)$ and $R(x, 6)$,find the values of $x$. Also,find the distances $QR$ and $PR$.

(X = ±4) Given that $Q(0, 1)$ is equidistant from $P(5, -3)$ and $R(x, 6)$,we have $PQ = QR$.
Using the distance formula $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$PQ^2 = QR^2$
$(5-0)^2 + (-3-1)^2 = (x-0)^2 + (6-1)^2$
$5^2 + (-4)^2 = x^2 + 5^2$
$25 + 16 = x^2 + 25$
$x^2 = 16 \implies x = \pm 4$.
Case $1$: If $R$ is $(4, 6)$:
$QR = \sqrt{(4-0)^2 + (6-1)^2} = \sqrt{16 + 25} = \sqrt{41}$.
$PR = \sqrt{(4-5)^2 + (6-(-3))^2} = \sqrt{(-1)^2 + 9^2} = \sqrt{1 + 81} = \sqrt{82}$.
Case $2$: If $R$ is $(-4, 6)$:
$QR = \sqrt{(-4-0)^2 + (6-1)^2} = \sqrt{16 + 25} = \sqrt{41}$.
$PR = \sqrt{(-4-5)^2 + (6-(-3))^2} = \sqrt{(-9)^2 + 9^2} = \sqrt{81 + 81} = \sqrt{162} = 9\sqrt{2}$.