Question

$\triangle abc$ is the image of $\triangle abc$ under a rotation about the origin, $(0,0)$. determine the angles of rotation. choose all answers that apply: a 90 clockwise b 90 counterclockwise c 180 d 270 clockwise e 270 counterclockwise

Explanation:

Step1: Recall rotation rules

• Rotation of \(90^{\circ}\) clockwise: \((x,y)\to(y, -x)\)
• Rotation of \(90^{\circ}\) counter - clockwise: \((x,y)\to(-y,x)\)
• Rotation of \(180^{\circ}\): \((x,y)\to(-x,-y)\)
• Rotation of \(270^{\circ}\) clockwise: \((x,y)\to(-y,x)\) (same as \(90^{\circ}\) counter - clockwise)
• Rotation of \(270^{\circ}\) counter - clockwise: \((x,y)\to(y, -x)\) (same as \(90^{\circ}\) clockwise)

Let's assume a point \(A(-4,2)\). After rotation, if we consider \(270^{\circ}\) clockwise (or \(90^{\circ}\) counter - clockwise) rotation:
Using the rule \((x,y)\to(-y,x)\), for \(A(-4,2)\), we get \(A'(- 2,-4)\)
Using the rule for \(90^{\circ}\) clockwise \((x,y)\to(y, -x)\), for \(A(-4,2)\) we get \(A'(2,4)\) (not matching)
Using the rule for \(180^{\circ}\) \((x,y)\to(-x,-y)\), for \(A(-4,2)\) we get \(A'(4,-2)\) (not matching)

If we consider the general property of rotation. A \(270^{\circ}\) clockwise rotation is equivalent to a \(90^{\circ}\) counter - clockwise rotation. Visually, when we rotate a figure \(270^{\circ}\) clockwise about the origin, it has the same result as rotating it \(90^{\circ}\) counter - clockwise.

Answer:

B. \(90\) counterclockwise, D. \(270\) clockwise