The coordinates of a point which is equidistant from the points (0, 0, 0), (o, 0, 0), (0, b, 0), (0,0, c) are given by

The coordinates of a point which is equidistant from the points (0, 0, 0), (o, 0, 0), (0, b, 0), (0,0, c) are given by

1. A

$\left(\frac{\mathrm{a}}{2},\frac{\mathrm{b}}{2},\frac{\mathrm{c}}{2}\right)$

2. B

$\left(-\frac{\mathrm{a}}{2},-\frac{\mathrm{b}}{2},\frac{\mathrm{c}}{2}\right)$

3. C

$\left(\frac{\mathrm{a}}{2},-\frac{\mathrm{b}}{2},-\frac{\mathrm{c}}{2}\right)$

4. D

$\left(-\frac{\mathrm{a}}{2},\frac{\mathrm{b}}{2},-\frac{\mathrm{c}}{2}\right)$

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Solution:

Let P be any point in a space. According to the given condition,

$\begin{array}{r}\sqrt{\left(\mathrm{x}-0{\right)}^{2}+\left(\mathrm{y}-0{\right)}^{2}+\left(\mathrm{z}-0{\right)}^{2}}=\sqrt{\left(\mathrm{x}-\mathrm{a}{\right)}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}}\\ =\sqrt{{\mathrm{x}}^{2}+\left(\mathrm{y}-\mathrm{b}{\right)}^{2}+{\mathrm{z}}^{2}}=\sqrt{{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+\left(\mathrm{z}-\mathrm{c}{\right)}^{2}}\end{array}$

On squaring both sides, we get

Similarly, we can solve the other values, we get

$\mathrm{y}=\frac{\mathrm{b}}{2},\mathrm{z}=\frac{\mathrm{c}}{2}$

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