To solve this problem, let's follow a step-by-step approach:
1. Identify the length of [tex]\(AB\)[/tex]:
The distance from point [tex]\(A\)[/tex] to point [tex]\(B\)[/tex] is 16 units.
2. Find the midpoint [tex]\(M\)[/tex]:
Point [tex]\(M\)[/tex] is the midpoint of [tex]\(AB\)[/tex], so [tex]\(M\)[/tex] is exactly halfway between [tex]\(A\)[/tex] and [tex]\(B\)[/tex].
[tex]\[
\text{Let } A = x \text{ and } B = x + 16
\][/tex]
[tex]\[
M = \frac{A + B}{2} = \frac{x + (x + 16)}{2} = \frac{2x + 16}{2} = x + 8
\][/tex]
3. Locate point [tex]\(K\)[/tex]:
Point [tex]\(K\)[/tex] is the midpoint of [tex]\(AM\)[/tex] and is given to be located at 2 on the number line.
[tex]\[
K = \frac{A + M}{2}
\][/tex]
Given [tex]\(K = 2\)[/tex],
[tex]\[
2 = \frac{A + M}{2} = \frac{x + (x + 8)}{2} = \frac{2x + 8}{2} = x + 4
\][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[
2 = x + 4
\][/tex]
[tex]\[
x = -2
\][/tex]
4. Determine points [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(M\)[/tex]:
Now that we have [tex]\(x = -2\)[/tex],
[tex]\[
A = -2
\][/tex]
[tex]\[
B = x + 16 = -2 + 16 = 14
\][/tex]
[tex]\[
M = x + 8 = -2 + 8 = 6
\][/tex]
5. Find the midpoint of [tex]\(AK\)[/tex]:
Knowing that [tex]\(K = 2\)[/tex] and [tex]\(A = -2\)[/tex],
[tex]\[
\text{Midpoint of } AK = \frac{A + K}{2} = \frac{-2 + 2}{2} = 0
\][/tex]
Thus, the largest possible midpoint of [tex]\(AK\)[/tex] is [tex]\(\boxed{0}\)[/tex].
