The coordinates of a point which is equidista | vedclass

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(A) Let the point be $P(x, y, z)$. Since $P$ is equidistant from the given points $O(0, 0, 0), A(a, 0, 0), B(0, b, 0)$,and $C(0, 0, c)$,we have $PO^2

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

Vedclass · Answer

(A) Let the point be $P(x, y, z)$. Since $P$ is equidistant from the given points $O(0, 0, 0), A(a, 0, 0), B(0, b, 0)$,and $C(0, 0, c)$,we have $PO^2 = PA^2 = PB^2 = PC^2$.$x^2 + y^2 + z^2 = (x - a)^2 + y^2 + z^2 = x^2 + (y - b)^2 + z^2 = x^2 + y^2 + (z - c)^2$.From $x^2 + y^2 + z^2 = (x - a)^2 + y^2 + z^2$,we get $x^2 = x^2 - 2ax + a^2$,which implies $2ax = a^2$,so $x = \frac{a}{2}$.Similarly,from $x^2 + y^2 + z^2 = x^2 + (y - b)^2 + z^2$,we get $y = \frac{b}{2}$.And from $x^2 + y^2 + z^2 = x^2 + y^2 + (z - c)^2$,we get $z = \frac{c}{2}$.Thus,the coordinates of the point are $\left( \frac{a}{2}, \frac{b}{2}, \frac{c}{2} \right)$.

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

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