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.

It is given that Q is equidistant from P and R. Using Distance Formula, we get
PQ = RQ

PQ=\(\sqrt{(5-0)^2 + (-3 - 1)^2} \)
\( = \sqrt{(-5)^2 + (-4)^2} \)
\( = \sqrt{25 + 16} \)
\( = \sqrt{41} \)
QR= \(\sqrt{(0 - x)^2 + (1 - 6)^2} \)
\( = \sqrt{(-x)^2 + (-5)^2} \)
\( = \sqrt{x^2 + 25} \)
\(\therefore \sqrt{41} = \sqrt{x^2 + 25} \)
On squaring both sides,
\(\Rightarrow 41 = x^2 + 25\)
\(\Rightarrow x^2 = 16\)
\(\Rightarrow x = \pm4\)
Coordinates of point R will be (4,6) or (-4,6),
If R is (4,6),
QR = \(\sqrt{(0 - 4)^2 + (1 - 6)^2} \)
\( = \sqrt{4^2 + 5^2} \)
\( = \sqrt{16 + 25} \)
\( = \sqrt{41} \)
PR = \(\sqrt{(5 - 4)^2 + (-3 - 6)^2} \)
\( = \sqrt{1^2 + 9^2} \)
\( = \sqrt{1 + 81} \)
\( = \sqrt{82} \)
If R is (-4,6),
QR = \(\sqrt{(0 + 4)^2 + (1 - 6)^2} \)
\( = \sqrt{4^2 + 5^2} \)
\( = \sqrt{16 + 25} \)
\( = \sqrt{41} \)
PR = \(\sqrt{(5 + 4)^2 + (-3 - 6)^2} \)
\( = \sqrt{9^2 + 9^2} \)
\( = \sqrt{81 + 81} \)
\( = 9\sqrt{2} \)