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The problem states that point \( I \) is located at \(-3\), and points \( J \) and \( K \) are each 9 units away from point \( I \).

To find the positions of \( J \) and \( K \) on the number line:

1. Point \( J \) is 9 units to the left of \( I \). Since \( I = -3 \), we subtract 9 from \(-3\):
   \[
   J = -3 - 9 = -12
   \]

2. Point \( K \) is 9 units to the right of \( I \). So, we add 9 to \(-3\):
   \[
   K = -3 + 9 = 6
   \]

Thus, \( J \) is located at \(-12\) and \( K \) is located at \(6\).

Would you like any further clarification or details?

Here are 5 related questions to explore:

1. What is the distance between points \( J \) and \( K \)?
2. How would you determine the midpoint between points \( J \) and \( K \)?
3. If point \( M \) is halfway between \( J \) and \( I \), where is it located?
4. What happens to the positions of \( J \) and \( K \) if point \( I \) moves to 0?
5. How would the distances change if \( J \) and \( K \) were 12 units away from point \( I \)?

**Tip**: When calculating distances on a number line, adding moves to the right and subtracting moves to the left.

Point I is located at -3. Points J and K are each 9 units away from point I. Plot point J and point K on the number line below. Use the answer blanks to state where J and K are located.

This problem involves locating points J and K, which are each 9 units away from point I located at -3, on a number line. Learn how to calculate the positions of J and K using simple addition and subtraction of integers.

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