Question

which rule describes the composition of transformations that maps \\( \triangle abc \\) to \\( \triangle abc \\)? \\( r _ { 0,90 ^ { \circ } } \circ r _ { x - \text { axis } } ( x, y ) \\) \\( r _ { x - \text { axis } } \circ r _ { 0,90 ^ { \circ } } ( x, y ) \\) \\( r _ { 0,180 ^ { \circ } } \circ r _ { x - \text { axis } } ( x, y ) \\) \\( r _ { x - \text { axis } } \circ r _ { 0,180 ^ { \circ } } ( x, y ) \\)

Explanation:

Step1: Analyze the order of transformations

First, consider the effect of each transformation. A rotation \(R_{0,90^{\circ}}\) rotates a point \((x,y)\) counter - clockwise by \(90^{\circ}\) around the origin, and the rule is \(R_{0,90^{\circ}}(x,y)=(-y,x)\). A reflection \(r_{x - axis}\) reflects a point \((x,y)\) over the \(x\) - axis, and the rule is \(r_{x - axis}(x,y)=(x,-y)\).

Let's assume a general point \((x,y)\) on \(\triangle ABC\).
If we first reflect over the \(x\) - axis: \(r_{x - axis}(x,y)=(x,-y)\). Then rotate the result \((x,-y)\) counter - clockwise by \(90^{\circ}\) around the origin. Using the rotation rule \(R_{0,90^{\circ}}(a,b)=(-b,a)\) (where \(a = x\) and \(b=-y\)), we get \(R_{0,90^{\circ}}(x,-y)=(y,x)\)

If we first rotate \((x,y)\) counter - clockwise by \(90^{\circ}\) around the origin: \(R_{0,90^{\circ}}(x,y)=(-y,x)\). Then reflect the result \((-y,x)\) over the \(x\) - axis. Using the reflection rule \(r_{x - axis}(a,b)=(a,-b)\) (where \(a=-y\) and \(b = x\)), we get \(r_{x - axis}(-y,x)=(-y,-x)\)

If we first reflect over the \(x\) - axis: \(r_{x - axis}(x,y)=(x,-y)\). Then rotate the result \((x,-y)\) counter - clockwise by \(180^{\circ}\) around the origin. Using the rotation rule \(R_{0,180^{\circ}}(a,b)=(-a,-b)\) (where \(a = x\) and \(b=-y\)), we get \(R_{0,180^{\circ}}(x,-y)=(-x,y)\)

If we first rotate \((x,y)\) counter - clockwise by \(180^{\circ}\) around the origin: \(R_{0,180^{\circ}}(x,y)=(-x,-y)\). Then reflect the result \((-x,-y)\) over the \(x\) - axis. Using the reflection rule \(r_{x - axis}(a,b)=(a,-b)\) (where \(a=-x\) and \(b=-y\)), we get \(r_{x - axis}(-x,-y)=(-x,y)\)

Let's take a specific point. Suppose \(A=(4,2)\)

• For \(R_{0,90^{\circ}}\circ r_{x - axis}(x,y)\): First \(r_{x - axis}(4,2)=(4,-2)\), then \(R_{0,90^{\circ}}(4,-2)=(2,4)\)
• For \(r_{x - axis}\circ R_{0,90^{\circ}}(x,y)\): First \(R_{0,90^{\circ}}(4,2)=(-2,4)\), then \(r_{x - axis}(-2,4)=(-2,-4)\)
• For \(R_{0,180^{\circ}}\circ r_{x - axis}(x,y)\): First \(r_{x - axis}(4,2)=(4,-2)\), then \(R_{0,180^{\circ}}(4,-2)=(-4,2)\)
• For \(r_{x - axis}\circ R_{0,180^{\circ}}(x,y)\): First \(R_{0,180^{\circ}}(4,2)=(-4,-2)\), then \(r_{x - axis}(-4,-2)=(-4,2)\)

Answer:

\(r_{x - axis}\circ R_{0,180^{\circ}}(x,y)\)