Question

find the coordinates of a point which is equidistant from the points (a, 0,0),(0,b,0),(0,0,c),(0,0,0)

Answer 1

this is the ans for your problem

Answer 2

$\fontsize{18}{10}{\textup{\textbf{The co-ordinates of the point are }}(\frac{a}{2},\frac{b}{2},\frac{c}{2}).$

Step-by-step explanation:

Let (x, y, z) represents the co-ordinates of the point that is equidistant from the points (a, 0, 0), (0, b, 0), (0, 0, c) and (0, 0, 0).

Now, (x, y, z) is equidistant from the points (a, 0, 0) and (0, 0, 0), so

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

Similarly, (x, y, z) is equidistant from the points (0, b, 0) and (0, 0, 0), so

$\sqrt{(x-0)^2+(y-b)^2+(z-0)^2}=\sqrt{(x-0)^2+(y-0)^2+(z-0)^2}\\\\\Rightarrow x^2+y^2-2by+b^2+z^2=x^2+y^2+z^2\\\\\Rightarrow b^2=2by\\\\\Rightarrow y=\dfrac{b}{2}.$

Again, (x, y, z) is equidistant from the points (0, 0, c) and (0, 0, 0), so

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

Thus, the required co-ordinates of the point are $(\frac{a}{2},\frac{b}{2},\frac{c}{2}).$

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