Question

which rule describes the composition of transformations that maps \\( \triangle abc \\) to \\( \triangle a ^ { prime } b ^ { prime } c ^ { prime } \\)?
\\( t _ { - 5, - 2 } circ r _ { x - \text { axis } } ( x, y ) \\)
\\( r _ { x - \text { axis } } circ t _ { - 5, - 2 } ( x, y ) \\)
\\( t _ { - 5, - 2 } circ r _ { 0,90 ^ { circ } } ( x, y ) \\)
\\( r _ { 0,90 ^ { circ } } circ t _ { - 5, - 2 } ( x, y ) \\)

Explanation:

Step1: Analyze the transformation order

First, consider the translation \(T_{- 6,-2}(x,y)=(x - 6,y - 2)\) and reflection \(r_{x - axis}(x,y)=(x,-y)\), rotation \(R_{0,90^{\circ}}(x,y)=(-y,x)\).
Let's assume a general point \((x,y)\) in \(\triangle ABC\).
If we first translate \(\triangle ABC\) using \(T_{-6,-2}\), a point \((x,y)\) becomes \((x - 6,y - 2)\). Then, if we reflect the translated figure over the \(x\) - axis, the \(y\) - coordinate changes its sign. So the transformation is \(r_{x - axis}\circ T_{-6,-2}(x,y)=r_{x - axis}(x - 6,y - 2)=(x - 6,-(y - 2))=(x - 6,-y + 2)\).
If we first reflect over the \(x\) - axis \(r_{x - axis}(x,y)=(x,-y)\) and then translate \(T_{-6,-2}(x,-y)=(x - 6,-y-2)\), which is not the correct transformation.
For rotation \(R_{0,90^{\circ}}(x,y)=(-y,x)\), if we first translate \(T_{-6,-2}(x,y)=(x - 6,y - 2)\) and then rotate \(R_{0,90^{\circ}}(x - 6,y - 2)=(-(y - 2),x - 6)=(-y + 2,x - 6)\). If we first rotate \(R_{0,90^{\circ}}(x,y)=(-y,x)\) and then translate \(T_{-6,-2}(-y,x)=(-y-6,x - 2)\), which is not correct.

Step2: Check the coordinates of vertices

Let's take a vertex of \(\triangle ABC\), say \(A(4,-3)\).
If we use \(r_{x - axis}\circ T_{-6,-2}\):
First, \(T_{-6,-2}(4,-3)=(4 - 6,-3 - 2)=(-2,-5)\). Then \(r_{x - axis}(-2,-5)=(-2,5)\).
If we use \(T_{-6,-2}\circ r_{x - axis}\):
First, \(r_{x - axis}(4,-3)=(4,3)\). Then \(T_{-6,-2}(4,3)=(4 - 6,3 - 2)=(-2,1)\) (incorrect).
For rotation - translation combinations:
Let \(A(4,-3)\). If we use \(T_{-6,-2}\circ R_{0,90^{\circ}}\):
First, \(R_{0,90^{\circ}}(4,-3)=(3,4)\). Then \(T_{-6,-2}(3,4)=(3 - 6,4 - 2)=(-3,2)\) (incorrect).
If we use \(R_{0,90^{\circ}}\circ T_{-6,-2}\):
First, \(T_{-6,-2}(4,-3)=(4 - 6,-3 - 2)=(-2,-5)\). Then \(R_{0,90^{\circ}}(-2,-5)=(5,-2)\) (incorrect).

Answer:

\(r_{x - axis}\circ T_{-6,-2}(x,y)\)