Question

which rotation is shown in the coordinate plane? 90° clockwise 180° counterclockwise 90° counterclockwise 180° clockwise

Explanation:

Step1: Recall rotation rules

For a \(90^{\circ}\) clockwise rotation \((x,y)\to(y, -x)\), for \(180^{\circ}\) rotation \((x,y)\to(-x,-y)\), for \(90^{\circ}\) counter - clockwise rotation \((x,y)\to(-y,x)\).

Step2: Assume a point

Let's take a point \(D(1, - 1)\) in the original triangle. After rotation, \(D'(-2,-1)\) is incorrect for \(90^{\circ}\) rotations. Let's check \(180^{\circ}\) rotation.
If we assume another point \(E(3, - 1)\), after \(180^{\circ}\) rotation \((x,y)\to(-x,-y)\), if we rotate \(E(3,-1)\) \(180^{\circ}\) counter - clockwise (or \(180^{\circ}\) clockwise as \(180^{\circ}\) clockwise and \(180^{\circ}\) counter - clockwise give the same result), we get \(E'(-3,1)\) (not in line). Wait, better to use the property of rotation.
The general property of \(180^{\circ}\) rotation: The rotation of a figure \(180^{\circ}\) about the origin maps \((x,y)\) to \((-x,-y)\). If we consider the orientation of the triangle. A \(90^{\circ}\) rotation (clockwise or counter - clockwise) would change the "slant" of the sides more distinctly. A \(180^{\circ}\) rotation (either clockwise or counter - clockwise) is equivalent as \((x,y)\to(-x,-y)\) for both. Looking at the orientation of the triangles (the direction of the vertices), a \(180^{\circ}\) rotation (since \(90^{\circ}\) rotations would make the triangle "point" in a different non - opposite direction compared to the original).

Answer:

\(180^{\circ}\) counterclockwise (or \(180^{\circ}\) clockwise as \(180^{\circ}\) clockwise and \(180^{\circ}\) counter - clockwise rotations are equivalent in terms of the final position of the figure in the coordinate plane)