Question

find a congruence transformation that maps (\triangle mbw) to (\triangle hxp).
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.
○ a. (t_{square square} circ r_{y - axis}(\triangle mbw))
○ b. (t_{square square} circ r_{(270^{circ},o)}(\triangle mbw))
○ c. (t_{square square} circ r_{(90^{circ},o)}(\triangle mbw))
○ d. (t_{square square} circ r_{x - axis}(\triangle mbw))

Explanation:

Step1: Analyze reflection over x - axis

First, consider the reflection over the \(x\) - axis. The rule for reflection over the \(x\) - axis is \((x,y)\to(x, - y)\).
Let's assume the coordinates of the vertices of \(\triangle MBW\). Suppose \(M(-6,6)\), \(B(-2,8)\), \(W(-4,4)\). After reflection over the \(x\) - axis, the coordinates become \(M'(-6,-6)\), \(B'(-2,-8)\), \(W'(-4,-4)\).

Step2: Analyze translation

Now, consider the translation. The rule for translation is \((x,y)\to(x + a,y + b)\).
If we translate the reflected points (after reflection over \(x\) - axis) by \(T_{\langle - 6,0
angle}\) (i.e., \(x=x-6\), \(y = y+0\)).
For \(M'(-6,-6)\): \((-6-6,-6) = (-12,-6)\) (incorrect).
Let's re - check.
Let's use another approach.
The coordinates of \(M(-6,6)\), \(B(-2,8)\), \(W(-4,4)\)
After reflection over \(x\) - axis: \(M_1(-6,-6)\), \(B_1(-2,-8)\), \(W_1(-4,-4)\)
If we consider the composition \(T_{\langle - 6,0
angle}\circ R_{x - axis}\)
\(R_{x - axis}(x,y)=(x,-y)\)
Let \(M(-6,6)\to M'(-6,-6)\), then \(T_{\langle - 6,0
angle}(x,y)=(x - 6,y)\)
\(M'(-6,-6)\to(-6-6,-6)=H(-12,-6)\)
\(B(-2,8)\to B'(-2,-8)\to(-2 - 6,-8)=X(-8,-8)\)
\(W(-4,4)\to W'(-4,-4)\to(-4-6,-4)=P(-10,-4)\)

Answer:

D. \(T_{\langle - 6,0
angle}\circ R_{x - axis}(\triangle MBW)\)