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# 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 = $$\sqrt{(5 + 4)^2 + (-3 - 6)^2}$$
$$= \sqrt{9^2 + 9^2}$$
$$= \sqrt{81 + 81}$$
$$= 9\sqrt{2}$$