First slide

Introduction to 3-D Geometry

Question

The point on z-axis which is equidistant from the points $(1, 5, 7)$ and $(5, 1, - 4)$

Easy

A

$(0, 0, \frac{3}{2})$
B

$(0, 0, 1)$
C

$(0, 0, \frac{13}{2})$
D

$(0, 0, 5)$

Solution

Suppose that $P (0, 0, z)$ is any point on $z -$ axis such that it is equidistant from the points $A (1, 5, 7)$ and $B (5, 1, - 4)$
It gives $P A = P B$
$\begin{matrix} {(z - 7)}^{2} = {(z + 4)}^{2} \\ - 14 z + 49 = 8 z + 16 \\ 22 z = 33 \\ z = \frac{3}{2} \end{matrix}$
Therefore, the point is $P (0, 0, \frac{3}{2})$

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