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

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Solution

PQ = QR(Given)
And distance between the points is given by
$\sqrt{{(x_{1} - x_{2})}_{1}^{2} + {(y_{1} - y_{2})}_{1}^{2}}$
So,
$\sqrt{(5 - 0)^{2} + (- 3 - 1)^{2}} = \sqrt{(0 - x)^{2} + (1 - 6)^{2}}$
$\sqrt{(5)^{2} + (- 4)^{2}} = \sqrt{(- x)^{2} + (- 5)^{2}}$
$\sqrt{25 + 16} = \sqrt{x^{2} + 25}$
$41 = x^{2} + 25$
$16 = x^{2}$
$x = \pm 4$
Therefore, point R is (4, 6) or ( - 4, 6).
When point R is (4, 6),
$P R = \sqrt{(5 - 4)^{2} + (- 3 - 6)^{2}} = \sqrt{1^{2} + (- 9)^{2}}$
$= \sqrt{1 + 81} = \sqrt{82}$
$Q R = \sqrt{(0 - 4)^{2} + (1 - 6)^{2}} = \sqrt{(- 4)^{2} + (- 5)^{2}}$
$= \sqrt{16 + 25} = \sqrt{41}$
When point R is (-4,6),
$P R = \sqrt{(5 - (- 4))^{2} + (- 3 - 6)^{2}} = \sqrt{(9)^{2} + (- 9)^{2}}$
$= \sqrt{81 + 81} = 9 \sqrt{2}$
$Q R = \sqrt{(0 - (- 4))^{2} + (1 - 6)^{2}} = \sqrt{(4)^{2} + (- 5)^{2}}$
$= \sqrt{16 + 25} = \sqrt{41}$

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