Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), which means PQ = QR

Step 1: Find the distance between PQ and QR using distance formula,

PQ = \(\sqrt{(5-0)^2 + (-3-1)^2}\)

= \(\sqrt{(-5)^2 + (-4)^2}\)

= \(\sqrt{25+16}\)

= \(\sqrt{41}\)

QR = \(\sqrt{(0-x)^2 + (1-6)^2}\)

= \(\sqrt{(-x)^2 + (-5)^2}\)

= \(\sqrt{x^2 + 25}\)

Step 2: Use PQ = QR

= \(\sqrt{41}\) = \(\sqrt{x^2 + 25}\)

Squaring both the sides, to omit square root

41 = x² + 25

x² = 16

x = ± 4

x = 4 or x = -4

Coordinates of Point R will be R (4, 6) or R (-4, 6),

If R (4, 6), then QR

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 (-4, 6), then

QR = \(\sqrt{(0+4)^2 + (1-6)^2}\)

= \(\sqrt{(4)^2 + (-5)^2}\)

= \(\sqrt{16+25}\)

= \(\sqrt{41}\)

PR = \(\sqrt{(5+4)^2 + (-3-6)^2}\)

= \(\sqrt{(9)^2 + (9)^2}\)

= \(\sqrt{81+81}\)

= \(9\sqrt{2}\)