Find the midpoint of the points (1802, 139, -4524) and (1885, 132, -4592).

Step-by-step solution

1. 1

   Use the three-dimensional midpoint formula. The midpoint of $A=(x_1,y_1,z_1)$ and $B=(x_2,y_2,z_2)$ averages each coordinate separately:

   $M=\left(\frac{x_1+x_2}{2},\;\frac{y_1+y_2}{2},\;\frac{z_1+z_2}{2}\right)$

   The three coordinates never interact, so this is three independent one-dimensional averages, not one three-dimensional computation.

2. 2

   Average the $x$-coordinates.

   $\frac{1802+1885}{2}=\frac{3687}{2}=1843.5$

   The sum is odd, so a half necessarily appears — that is not a rounding artefact.

3. 3

   Average the $y$-coordinates.

   $\frac{139+132}{2}=\frac{271}{2}=135.5$

4. 4

   Average the $z$-coordinates, minding the negatives.

   $\frac{-4524+(-4592)}{2}=\frac{-9116}{2}=-4558$

   Both values are negative, so their sum is more negative than either; a common slip is subtracting instead of adding here.

5. 5

   State the midpoint and check it lies between the endpoints.

   $M=(1843.5,\;135.5,\;-4558)$

   Each coordinate of $M$ falls strictly between the corresponding coordinates of $A$ and $B$: $1802<1843.5<1885$, $132<135.5<139$, and $-4592<-4558<-4524\;\checkmark$. A stronger check is that $M-A=(41.5,-3.5,-34)$ equals $B-M=(41.5,-3.5,-34)$ exactly, so $M$ really is equidistant along the segment.