Question

Difficulty: MediumIntegers, Absolute Value, and Number Lines

On a standard number line, point AA has coordinate 28-28 and point BB has coordinate 1212. Point CC is the midpoint of segment ABAB. Point DD is positioned on the line such that point BB is the midpoint of segment CDCD. What is the distance between point CC and point DD on the number line?

Answer: 40

Answer

The distance between point CC and point DD on the number line is 40.
The midpoint CC of segment ABAB is found by averaging the coordinates: 28+122=8\frac{-28 + 12}{2} = -8. Because point B(12)B(12) is the midpoint of segment CDCD, set up the equation 8+D2=12\frac{-8 + D}{2} = 12, yielding D=32D = 32. The distance between point C(8)C(-8) and point D(32)D(32) is the absolute value of their difference: 32(8)=40|32 - (-8)| = 40.

Step-by-Step Solution

1
Calculate the coordinate of point CC, which is the midpoint of segment ABAB.
The coordinate of point CC is 8-8.
The midpoint of two points on a number line is given by their average: 28+122=8\frac{-28 + 12}{2} = -8.
2
Determine the coordinate of point DD using the midpoint relationship for segment CDCD.
The coordinate of point DD is 3232.
Since point B(12)B(12) is the midpoint of segment CDCD, C+D2=12    8+D2=12\frac{C + D}{2} = 12 \implies \frac{-8 + D}{2} = 12, which solves to D=32D = 32.
3
Find the distance between point CC and point DD.
The distance is 4040.
Distance on a number line is the absolute value of the difference between coordinates: 32(8)=40=40|32 - (-8)| = |40| = 40.

Key Concept

Midpoint and Absolute Value Distance on a Number Line

Alternative Method

Alternatively, note that the distance of segment ABAB is 12(28)=40|12 - (-28)| = 40. Since CC is the midpoint of ABAB, the length of segment CBCB is 402=20\frac{40}{2} = 20. Since BB is the midpoint of segment CDCD, the length of segment BDBD must equal the length of segment CBCB, which is 2020. Therefore, the total distance from CC to DD is CB+BD=20+20=40CB + BD = 20 + 20 = 40.
Estimated Time:1m 15s
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