Question

Determine the point on z-axis which is equidistant from the points (1, 5, 7) and (5, 1, –4).

Answer 1

Let M be the point on the z-axis.
Then, the coordinates of M will be\[\left( 0, 0, z \right)\] 

Let M be equidistant from the points A\[\left( 1, 5, 7 \right)\]and B \[\left( 5, 1, - 4 \right)\]

AM =\[\sqrt{\left( 0 - 1 \right)^2 + \left( 0 - 5 \right)^2 + \left( z - 7 \right)^2}\]

\[= \sqrt{\left( - 1 \right)^2 + \left( - 5 \right)^2 + \left( z - 7 \right)^2}\]
\[ = \sqrt{1 + 25 + z^2 - 14z + 49}\]
\[ = \sqrt{z^2 - 14z + 75}\]

BM =\[\sqrt{\left( 0 - 5 \right)^2 + \left( 0 - 1 \right)^2 + \left( z + 4 \right)^2}\]

\[= \sqrt{\left( - 5 \right)^2 + \left( - 1 \right)^2 + z^2 + 8z + 16}\]
\[ = \sqrt{25 + 1 + z^2 + 8z + 16}\]
\[ = \sqrt{z^2 + 8z + 42}\]

Then, AM = BM

\[\therefore \sqrt{z^2 - 14z + 75} = \sqrt{z^2 + 8z + 42}\]
\[ z^2 - 14z + 75 = z^2 + 8z + 42\]
\[ - 14z - 8z = 42 - 75\]
\[ - 22z = - 33\]
\[z = \frac{33}{22}\]
\[z = \frac{3}{2}\]

Thus, the coordinates of M are \[\left( 0, 0, \frac{3}{2} \right)\]