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}, 0)}(\triangle byw) \\)
\\( \bigcirc \\) c. \\( t \square \square \circ r_{x - \text { axis }}(\triangle byw) \\)
\\( \bigcirc \\) d. \\( t \square \square \circ r_{(90^{\circ}, 0)}(\triangle byw) \\)

Explanation:

Step1: Reflect over the \(x -\)axis

Reflecting \(\triangle BYW\) over the \(x -\)axis. The rule for reflection over the \(x -\)axis is \((x,y)\to(x, - y)\).
For example, if \(B(-6,5)\), after reflection over the \(x -\)axis, it becomes \(B'(-6,-5)\).

Step2: Translate

After reflection over the \(x -\)axis, we need to translate.
The \(x -\)coordinate of the reflected point (after \(R_{x - axis}\)): Let's assume a general point \((x,y)\) after \(R_{x - axis}\) is \((x,-y)\).
We observe that we need to move the reflected triangle \(4\) units to the right. The translation rule \(T_{(a,b)}\) is \((x,y)\to(x + a,y + b)\). Here \(a = 4,b=0\).

Answer:

C. \(T_{(4,0)}\circ R_{x - axis}(\triangle BYW)\)