Question

analyze when a rotation and a reflection are performed as a composition of transformations on a figure, does the order of the transformations sometimes, always, or never affect the location of the final image? justify your argument. always ; the order of rotating by 180° about the origin and reflecting in the line y = x does change the location of the final image.

Explanation:

• Composition of transformations: A rotation and a reflection are both rigid transformations. But their order can affect the final result.
• Example: Consider a point \((x,y)\).
• First rotate by \(180^{\circ}\) about the origin: The rule for a \(180^{\circ}\) rotation about the origin is \((x,y)\to(-x,-y)\). Then reflect in the line \(y = x\): The rule for reflection in the line \(y=x\) is \((a,b)\to(b,a)\). So \((-x,-y)\to(-y,-x)\).
• First reflect in the line \(y = x\): \((x,y)\to(y,x)\). Then rotate by \(180^{\circ}\) about the origin: \((y,x)\to(-y,-x)\). In this case, the order does not change the final image.
• Consider another example: Let's take a point \((1,0)\).
• First rotate by \(90^{\circ}\) counter - clockwise about the origin: The rule for a \(90^{\circ}\) counter - clockwise rotation about the origin is \((x,y)\to(-y,x)\). So \((1,0)\to(0,1)\). Then reflect in the \(x\) - axis: The rule for reflection in the \(x\) - axis is \((a,b)\to(a,-b)\). So \((0,1)\to(0, - 1)\).
• First reflect in the \(x\) - axis: \((1,0)\to(1,0)\). Then rotate by \(90^{\circ}\) counter - clockwise about the origin: \((1,0)\to(0,1)\). Here the order changes the final image.

Answer:

Sometimes; because there are cases (e.g., \(180^{\circ}\) rotation and reflection in \(y = x\)) where the order does not change the final image and cases (e.g., \(90^{\circ}\) rotation and \(x\) - axis reflection) where the order changes the final image.