In the figure below, B lies between A and C.Find the location of B so that AB is 16 of AC.A−27C3B?

1. Break Down the Problem

We need to find the position of point B such that the distance AB is one-sixth of the distance AC. The points A, B, and C are collinear, with B located between A and C.

2. Relevant Concepts

Let:

• Distance $AC = d$
• Distance $AB = \frac{1}{6}AC = \frac{1}{6}d$
• Consequently, distance $BC = AC - AB = d - \frac{1}{6}d = \frac{5}{6}d$

3. Analysis and Detail

1. Let distance $AC = d$.

2. Since $AB = \frac{1}{6}d$, we can express the total distances as:

   • $AB + BC = AC$
   • $\frac{1}{6}d + BC = d$

3. Expressing $BC$:

   • $BC = d - \frac{1}{6}d = \frac{5}{6}d$

4. Therefore, we know that B divides the line segment AC into two parts: $AB$ and $BC$.

4. Verify and Summarize

We've correctly established that since B divides AC, its distance from A is one-sixth the distance from A to C, while its distance from C is five-sixths of the distance from A to C.

Final Answer

The location of B can be expressed in terms of the total distance AC as follows:

• $AB = \frac{1}{6}AC$
• $BC = \frac{5}{6}AC$

Thus, point B is located at one-sixth of the way from A to C on the line segment.