If Q (0,1) is equidistant from P(5, – 3) and R(x, 6),

Solution:

Here, $Q P=\sqrt{(5-0)^{2}+\mid(-3)-11^{2}}=\sqrt{5^{2}+(-4)^{2}}$

$=\sqrt{25+16}=\sqrt{41}$

$Q R=\sqrt{(x-0)^{2}+(6-1)^{2}}=\sqrt{x^{2}+5^{2}}=\sqrt{x^{2}+25}$

$\because \quad \mathrm{QP}=\mathrm{QR}$

$\therefore \quad \sqrt{41}=\sqrt{x^{2}+25}$

Squaring both sides, we have $x^{2}+25=41$

$\Rightarrow x^{2}+25-41=0$

$\Rightarrow x^{2}-16=0 \Rightarrow x=\pm \sqrt{\mathbf{1 6}}=\pm 4$

Thus, the point R is (4, 6) or (–4, 6)

Now,

$Q R=\sqrt{((+4)-(0))^{2}+(6-1)^{2}}=\sqrt{16+25}=\sqrt{41}$

and $P R=\sqrt{(+4-5)^{2}+(6+3)^{2}}$

$\Rightarrow P R=\sqrt{(-4-5)^{2}+(6+3)^{2}}$

or $\sqrt{(4-5)^{2}+(6+3)^{2}}$

$\Rightarrow P R=\sqrt{(-9)^{2}+9^{2}}$ or $\sqrt{\mathbf{1}+\mathbf{8 1}}$

$\Rightarrow P R=\sqrt{\mathbf{2} \times \mathbf{9}^{2}}$ or $\sqrt{\mathbf{8 2}}$

$\Rightarrow P R=\mathbf{9} \sqrt{\mathbf{2}}$ or $\sqrt{\mathbf{8 2}}$