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.