Question

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

• A. $(\frac{a}{2},\frac{b}{2},\frac{c}{2})$
• B. $(\frac{-a}{2},\frac{-b}{2},\frac{c}{2})$
• C. $(\frac{a}{2},\frac{-b}{2},\frac{-c}{2})$
• D. $(\frac{-a}{2},\frac{b}{2},\frac{-c}{2})$

Answer 1

The correct option is A

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

Let that point be P$(x,y,z)$

Given points A$(0,0,0),B(a,0,0),C(0,b,0)andD(0,0,c)$

$AP=BP=CP=DP$

$A{P}^2=B{P}^2=C{P}^2=D{P}^2$

$\stackrel{}{A{P}^2=B{P}^2}$

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

$\Rightarrow x^2=x^2+a^2-2ax$

$x=\frac{a}{2}$

similarly,

$A{P}^2=C{P}^2$

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

$y^2=y^2+b^2-2by$

$y=\frac{b}{2}$

and $A{P}^2=D{P}^2$

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

$z^2=z^2+c^2-2zc$

z = $\frac{c}{2}$

$\Rightarrow$ coordinate of the point C is $(\frac{a}{2},\frac{b}{2},\frac{c}{2})$