For $\# 2$ - 3, draw the image of the figure under the given rotation. 2. $\triangle P Q R ; 90^{\circ}$ about the origin. When no direction is specified, you can assume the rotation is counterclockwise Transformation Rule: $(x, y) \rightarrow$ \begin{tabular}{|c|c|} \hline $\begin{array}{c}\text { Preimage Coordinates } \\ (x, y)\end{array}$ & Image \\ \hline & \\ \hline & \\ \hline & \\ \hline \end{tabular}

Step 1 ：The problem is asking to rotate a triangle 90 degrees counterclockwise about the origin. The transformation rule for a 90 degree counterclockwise rotation about the origin is \((x, y) \rightarrow (-y, x)\). This means that for each point \((x, y)\) in the triangle, we replace x with -y and y with x to get the new coordinates of the point after rotation.

Step 2 ：Let's apply this rule to the vertices of the triangle \(PQR\) with coordinates \((1, 2)\), \((3, 4)\), and \((5, 6)\).

Step 3 ：For point P \((1, 2)\), replacing x with -y and y with x gives us \((-2, 1)\).

Step 4 ：For point Q \((3, 4)\), replacing x with -y and y with x gives us \((-4, 3)\).

Step 5 ：For point R \((5, 6)\), replacing x with -y and y with x gives us \((-6, 5)\).

Step 6 ：Thus, the image of the triangle \(PQR\) under a \(90^\circ\) rotation about the origin is the triangle with vertices at \((-2, 1)\), \((-4, 3)\), and \((-6, 5)\).

Step 7 ：In other words, \(\triangle PQR\) is transformed to \(\triangle P'Q'R'\) where \(P' = (-2, 1)\), \(Q' = (-4, 3)\), and \(R' = (-6, 5)\).

Step 8 ：\(\boxed{\text{Final Answer: } P' = (-2, 1), Q' = (-4, 3), R' = (-6, 5)}\)