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# Find the coordinates of the point which is equidistant from the points

Question:

Find the coordinates of the point which is equidistant from the points A(a, 0, 0), B(0, b, 0), C(0, 0, c) and O(0, 0, 0).

Solution:

Consider, D(x,y,z) point equidistant from points A(a, 0, 0), B(0, b, 0), C(0, 0, c) and O(0, 0, 0).

$\sqrt{(x-a)^{2}+(y-0)^{2}+(z-0)^{2}}=\sqrt{(x-0)^{2}+(y-0)^{2}+(z-0)^{2}}$

Squaring both sides,

$(x-a)^{2}+(y-0)^{2}+(z-0)^{2}=(x-0)^{2}+(y-0)^{2}+(z-0)^{2}$

$x^{2}+2 a x+a^{2}+y^{2}+z^{2}=x^{2}+y^{2}+z^{2}$

$a(2 x-a)=0$

as $a \neq 0$

$x=a / 2$

∴ BD = 0D

$\sqrt{(x-a)^{2}+(y-0)^{2}+(z-0)^{2}}=\sqrt{(x-0)^{2}+(y-0)^{2}+(z-0)^{2}}$

Squaring both sides,

$(x-0)^{2}+(y-b)^{2}+(z-0)^{2}=(x-0)^{2}+(y-0)^{2}+(z-0)^{2}$

$x^{2}+y^{2}+2 b y+b^{2}+z^{2}=x^{2}+y^{2}+z^{2}$

$b(2 y-b)=0$

as $b \neq 0$

$y=b / 2$

$\therefore C D=0 D$

$\sqrt{(x-0)^{2}+(y-0)^{2}+(z-c)^{2}}=\sqrt{(x-0)^{2}+(y-0)^{2}+(z-0)^{2}}$

Squaring both sides,

$(x-0)^{2}+(y-0)^{2}+(z-c)^{2}=(x-0)^{2}+(y-0)^{2}+(z-0)^{2}$

$x^{2}+y^{2}+z^{2}+2 c z+c^{2}=x^{2}+y^{2}+z^{2}$

$c(2 z-c)=0$

as $c \neq 0 .$

$z=c / 2$

Therefore, the pint $\mathrm{D}(\mathrm{a} / 2, \mathrm{~b} / 2, \mathrm{c} / 2)$ is equidistant to points $\mathrm{A}(\mathrm{a}, 0,0), \mathrm{B}(0, \mathrm{~b}, 0), \mathrm{C}(0,0, \mathrm{c})$ and $O(0,0,0)$.