On a certain number line, [tex]AB[/tex] is 16 units long. Point [tex]M[/tex] is the midpoint of [tex]AB[/tex]. Point [tex]K[/tex] is the midpoint of [tex]AM[/tex] and is located at 2 on the number line. What is the largest possible midpoint of [tex]AK[/tex]?

A. 0
B. 4
C. 6
D. 8



Answer :

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].