Question

find a congruence transformation that maps \\( \triangle byw \\) to \\( \triangle vng \\).
which congruence transformation maps the preimage to the congruent image? select the correct choice below and fill in the answer boxes to complete you choice.
\\( \bigcirc \\) a. \\( t \square \square \circ r_{y - \text{axis}}(\triangle byw) \\)
\\( \bigcirc \\) b. \\( t \square \square \circ r_{(270^{\circ},o)}(\triangle byw) \\)
\\( \bigcirc \\) c. \\( t \square \square \circ r_{x - \text{axis}}(\triangle byw) \\)
\\( \bigcirc \\) d. \\( t \square \square \circ r_{(90^{\circ},o)}(\triangle byw) \\)

Explanation:

Step1: Analyze the reflection

First, consider the reflection. Reflecting \(\triangle BYW\) over the \(x -\)axis. The rule for reflecting a point \((x,y)\) over the \(x -\)axis is \((x,y)\to(x, - y)\).

Step2: Analyze the translation

After reflection over the \(x -\)axis, we need to translate. Let's assume a general translation rule \(T_{(a,b)}\) where \(T_{(a,b)}(x,y)=(x + a,y + b)\).
Take a point, say \(B(-7,4)\). After reflection over the \(x -\)axis, it becomes \((-7,-4)\). The corresponding point \(V(-9,-4)\). To get from \((-7,-4)\) to \((-9,-4)\), we use the translation \(T_{(- 2,0)}\) (since \(x=-7+a=-9\) gives \(a=-2\) and \(y=-4 + b=-4\) gives \(b = 0\)).

Check another point: \(Y(5,6)\). After reflection over the \(x -\)axis, it is \((5,-6)\). After translation \(T_{(-2,0)}\), it is \((5-2,-6)=(3,-6)\). But \(N(3,-6)\) (if we consider the congruence). Also, \(W(-1,3)\) after reflection over \(x -\)axis is \((-1,-3)\), after translation \(T_{(-2,0)}\) is \((-3,-3)\) and \(G(-3,-1)\) (wait, no, let's re - check the order.

Let's use the formula for composition of transformations. If we first reflect \(\triangle BYW\) over the \(x -\)axis:
The coordinates of \(B(-7,4)\to B'(-7,-4)\), \(Y(5,6)\to Y'(5,-6)\), \(W(-1,3)\to W'(-1,-3)\)
Then translate \(T_{(-2,0)}\): \(B'(-7,-4)\to(-7-2,-4)=V(-9,-4)\), \(Y'(5,-6)\to(5 - 2,-6)=N(3,-6)\), \(W'(-1,-3)\to(-1-2,-3)=G(-3,-3)\) (typo in the problem statement's figure labels maybe, but the transformation \(T_{(-2,0)}\circ R_{x - axis}\) works.

Answer:

C. \(T_{(-2,0)}\circ R_{x - axis}(\triangle BYW)\)