Question

On a number line, point A is located at 1, point C is located at −4, and point B lies between points A and C. What is the location of B such that the ratio of CB:BA is 3:1?

−2.75
−2.3
−0.25
1.25

Answer 1

To find the location of point B such that the ratio of the distances \( CB:BA \) is 3:1, we can first denote the positions of points A, B, and C:

- Point A is at \( 1 \)
- Point C is at \( -4 \)
- Point B is at \( x \) (the unknown we need to determine)

Next, we can express the distances \( CB \) and \( BA \):

- The distance \( CB \) (from C to B) is calculated as \( |x - (-4)| = |x + 4| \).
- The distance \( BA \) (from B to A) is calculated as \( |x - 1| \).

Given the ratio \( CB:BA = 3:1 \), we can express this as:
\[
\frac{|x + 4|}{|x - 1|} = 3
\]

Cross-multiplying gives:
\[
|x + 4| = 3|x - 1|
\]

Next, we consider two cases based on the absolute values.

### Case 1: \( x + 4 \geq 0 \) and \( x - 1 \geq 0 \)

In this case, \( |x + 4| = x + 4 \) and \( |x - 1| = x - 1 \). The equation simplifies to:
\[
x + 4 = 3(x - 1)
\]
\[
x + 4 = 3x - 3
\]
\[
4 + 3 = 3x - x
\]
\[
7 = 2x
\]
\[
x = 3.5
\]

Since \( 3.5 \) is not between \( -4 \) and \( 1 \), we discard this case.

### Case 2: \( x + 4 \geq 0 \) and \( x - 1 < 0 \)

In this case, \( |x + 4| = x + 4 \) and \( |x - 1| = - (x - 1) = -x + 1 \). The equation simplifies to:
\[
x + 4 = 3(-x + 1)
\]
\[
x + 4 = -3x + 3
\]
\[
x + 3x = 3 - 4
\]
\[
4x = -1
\]
\[
x = -\frac{1}{4} \text{ or } -0.25
\]

This value is between \( -4 \) and \( 1 \).

### Case 3: \( x + 4 < 0 \) and \( x - 1 < 0 \)

In this case, \( |x + 4| = - (x + 4) \) and \( |x - 1| = - (x - 1) = -x + 1 \). Thus:
\[
-(x + 4) = 3(-x + 1)
\]
\[
-x - 4 = -3x + 3
\]
\[
-4 - 3 = -3x + x
\]
\[
-7 = -2x
\]
\[
x = \frac{7}{2} = 3.5
\]

Again, this value is not between \( -4 \) and \( 1 \).

The only valid solution derived from the cases is:
\[
x = -0.25
\]

Thus, point B is located at \( -0.25 \).

The answer is:
\[
\boxed{-0.25}
\]