To rotate the triangle $\triangle PQR$ counterclockwise by 270° around the origin, we apply the following transformation to each of the points $P(x_1, y_1)$, $Q(x_2, y_2)$, and $R(x_3, y_3)$:

• The rule for a 270° counterclockwise rotation about the origin is $(x, y) \rightarrow (y, -x)$.

Let's identify the coordinates of points $P$, $Q$, and $R$ from the graph, apply the transformation, and find the new positions.

1. Identify the coordinates:

   • $P$ seems to be at $(10, 15)$.
   • $Q$ seems to be at $(5, 10)$.
   • $R$ seems to be at $(10, 0)$.

2. Apply the 270° rotation rule $(x, y) \rightarrow (y, -x)$:

   • For $P(10, 15)$, the new coordinates will be $(15, -10)$.
   • For $Q(5, 10)$, the new coordinates will be $(10, -5)$.
   • For $R(10, 0)$, the new coordinates will be $(0, -10)$.

So, the rotated triangle $\triangle P'Q'R'$ will have the following coordinates:

• $P'(15, -10)$,
• $Q'(10, -5)$,
• $R'(0, -10)$.

Would you like further clarification on this, or any specific details on how this rotation works?

Here are some related questions:

1. What are the steps for performing a 90° counterclockwise rotation?
2. How would you rotate a point 180° about the origin?
3. What is the formula for rotating a point clockwise by 270°?
4. How would a reflection across the x-axis affect the points of a triangle?
5. How does a translation differ from a rotation in terms of transformation?

Tip: A 270° counterclockwise rotation is the same as a 90° clockwise rotation, which can simplify some problems.