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In this problem, you are working with translations on a coordinate plane. The translation vector given is \( \langle 1, 3 \rangle \), which means that each point on the pre-image needs to be translated by moving 1 unit to the right (in the \(x\)-direction) and 3 units up (in the \(y\)-direction).

The pre-image coordinates are:
- \( P(2, -5) \)
- \( Q(6, -9) \)
- \( R(3, 1) \)

To find the image coordinates (P', Q', R'), you will apply the translation vector \( \langle 1, 3 \rangle \) to each point:

### 1. For point P(2, -5):
   \[
   P' = (2 + 1, -5 + 3) = (3, -2)
   \]

### 2. For point Q(6, -9):
   \[
   Q' = (6 + 1, -9 + 3) = (7, -6)
   \]

### 3. For point R(3, 1):
   \[
   R' = (3 + 1, 1 + 3) = (4, 4)
   \]

Thus, the correct image coordinates are:
- \( P' = (3, -2) \)
- \( Q' = (7, -6) \)
- \( R' = (4, 4) \)

Let me know if you need more details or have any questions!

Here are 5 related questions for further practice:
1. What is the general formula for translating a point by a vector \( \langle a, b \rangle \)?
2. How would you translate a point by \( \langle -2, 4 \rangle \)?
3. What happens to the coordinates if you translate by \( \langle 0, 0 \rangle \)?
4. Can a translation ever change the shape of a figure?
5. How can you determine if two figures are congruent after a translation?

**Tip**: When translating, always add the translation vector components directly to the corresponding \(x\) and \(y\) coordinates.

Translate the pre-image points P(2, -5), Q(6, -9), and R(3, 1) using the translation vector ⟨1, 3⟩ and match the image coordinates.

Learn how to translate the points P(2, -5), Q(6, -9), and R(3, 1) using the vector ⟨1, 3⟩. This exercise covers essential concepts in coordinate geometry, suitable for students in Grades 6-8.

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