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The image you uploaded shows a graph from a Khan Academy exercise. The problem asks for the area of a triangle \( \triangle PQR \) with vertices \( P \), \( Q \), and \( R \) located at points \( (1,1) \), \( (2,5) \), and \( (-1,2) \) respectively.

### Steps to Solve:

1. **Identify the vertices:**
   - \( P(1, 1) \)
   - \( Q(2, 5) \)
   - \( R(-1, 2) \)

2. **Use the area formula for a triangle given its vertices:**
   The area \( A \) of a triangle given vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is:

   \[
   A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
   \]

3. **Substitute the coordinates of the points \( P(1, 1) \), \( Q(2, 5) \), and \( R(-1, 2) \) into the formula:**

   \[
   A = \frac{1}{2} \left| 1(5 - 2) + 2(2 - 1) + (-1)(1 - 5) \right|
   \]

4. **Simplify the expression:**

   \[
   A = \frac{1}{2} \left| 1(3) + 2(1) + (-1)(-4) \right|
   \]
   \[
   A = \frac{1}{2} \left| 3 + 2 + 4 \right|
   \]
   \[
   A = \frac{1}{2} \times 9 = \frac{9}{2} = 4.5
   \]

So, the area of triangle \( \triangle PQR \) is **4.5 square units**.

Would you like further details or have any questions? Here are some additional related questions:

1. How would the area change if one of the vertices is shifted horizontally or vertically?
2. Can you explain why the formula for the area of a triangle works?
3. What happens to the area if all three points are collinear?
4. How can you find the centroid of triangle \( \triangle PQR \)?
5. What would be the area if one of the vertices is at the origin?

**Tip:** To ensure accuracy in geometry problems, always double-check the coordinates and calculations before finalizing your answer.

What is the area of the triangle PQR with vertices (1,1), (2,5), and (-1,2)?

This page explains how to calculate the area of triangle PQR with given vertices at (1,1), (2,5), and (-1,2) using the formula for the area of a triangle in a coordinate plane. Suitable for students in Grades 8-10.

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