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$(0,0,0),(\mathrm{a}, 0,0),(0, \mathrm{~b}, 0)$ and $(0,0, \mathrm{c})$ are four distinct points. What are the coordinates of the point which is equidistant from the four points?
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Verified Answer
The correct answer is:
$\left(\frac{a}{2}, \frac{b}{2}, \frac{c}{2}\right)$
$\mathrm{A}(0,0,0), \mathrm{B}(\mathrm{a}, 0,0), \mathrm{C}(0, \mathrm{~b}, 0), \mathrm{D}(0,0, \mathrm{c})$
Let the equidistant point be $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})$
i.e, $\mathrm{AP}=\mathrm{BP}, \mathrm{AP}=\mathrm{CP}, \mathrm{AP}=\mathrm{DP}$
$\Rightarrow \sqrt{(x-0)^{2}+(y-0)^{2}+(z-0)^{2}}$
$=\sqrt{(x-a)^{2}+y^{2}+z^{2}}$
$\Rightarrow x^{2}+y^{\prime}+z^{\prime}=(x-a)^{2}+y^{\prime}+z^{\prime}$
$\Rightarrow x^{2}=x^{2}-2 a x+a^{2}$
$\Rightarrow a^{2}-2 a x=0$
$\Rightarrow a(a-2 x)=0$
Since $a \neq 0, a=2 x \Rightarrow x=\frac{a}{2}$
Similarly, we will get $\mathrm{y}=\frac{\mathrm{b}}{2}, \mathrm{z}=\frac{\mathrm{c}}{2}$
Let the equidistant point be $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})$
i.e, $\mathrm{AP}=\mathrm{BP}, \mathrm{AP}=\mathrm{CP}, \mathrm{AP}=\mathrm{DP}$
$\Rightarrow \sqrt{(x-0)^{2}+(y-0)^{2}+(z-0)^{2}}$
$=\sqrt{(x-a)^{2}+y^{2}+z^{2}}$
$\Rightarrow x^{2}+y^{\prime}+z^{\prime}=(x-a)^{2}+y^{\prime}+z^{\prime}$
$\Rightarrow x^{2}=x^{2}-2 a x+a^{2}$
$\Rightarrow a^{2}-2 a x=0$
$\Rightarrow a(a-2 x)=0$
Since $a \neq 0, a=2 x \Rightarrow x=\frac{a}{2}$
Similarly, we will get $\mathrm{y}=\frac{\mathrm{b}}{2}, \mathrm{z}=\frac{\mathrm{c}}{2}$
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