Finding Distance Between Points C and H on a Number Line: From -3 to 2

Distance Calculation with Negative Coordinates

What is the distance between C and H?

AAA-5-5-5BBB-4-4-4CCC-3-3-3DDD-2-2-2EEE-1-1-1FFF000GGG111HHH222III333JJJ444

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1

Understand the problem

What is the distance between C and H?

AAA-5-5-5BBB-4-4-4CCC-3-3-3DDD-2-2-2EEE-1-1-1FFF000GGG111HHH222III333JJJ444

2

Step-by-step solution

We first mark the letter C on the number line and then proceed towards the letter H:

AAA-5-5-5BBB-4-4-4CCC-3-3-3DDD-2-2-2EEE-1-1-1FFF000GGG111HHH222III333JJJ444

Note that the distance between the two letters is 5 steps.

3

Final Answer

5

Key Points to Remember

Essential concepts to master this topic
  • Distance Formula: Count units between points or use absolute value method
  • Technique: From C(-3) to H(2): |2 - (-3)| = |5| = 5
  • Check: Count spaces on number line: -3 → -2 → -1 → 0 → 1 → 2 = 5 steps ✓

Common Mistakes

Avoid these frequent errors
  • Subtracting coordinates without absolute value
    Don't just calculate 2 - (-3) = 5 and think you're done! Without absolute value, you might get negative distances when coordinates are reversed. Always use |final - initial| or count actual units on the number line.

Practice Quiz

Test your knowledge with interactive questions

\( 5 < -5 \)

AAAKKK-5-5-5BBB-4-4-4CCC-3-3-3DDD-2-2-2EEE-1-1-1FFF000GGG111HHH222III333JJJ444555

FAQ

Everything you need to know about this question

Why do I need absolute value when finding distance?

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Distance is always positive - you can't have negative distance! The absolute value ensures your answer is positive regardless of which point you subtract from which.

What's the easiest way to find distance on a number line?

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Use the formula distance = |point₂ - point₁|. For C(-3) and H(2): distance = |2 - (-3)| = |5| = 5 units.

Can I just count the spaces between points?

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Yes! Counting is a great visual method. From -3 to 2, count each unit: -3→-2→-1→0→1→2 gives you 5 spaces.

Does it matter which point I start from?

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No! Distance from C to H equals distance from H to C. That's why we use absolute value: |2-(-3)| = |(-3)-2| = 5.

What if both points are negative?

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Same process! For points at -5 and -2: distance = |(-2) - (-5)| = |3| = 3 units. The absolute value handles all cases.

How do I avoid calculation errors with negative numbers?

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Remember: subtracting a negative means adding. So 2 - (-3) becomes 2 + 3 = 5. Then apply absolute value if needed.

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