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1 point

[tex]$\triangle XYZ$[/tex] is reflected across the [tex]$x$[/tex]-axis and then dilated by a scale factor of 2, producing [tex]$\triangle X' Y' Z'$[/tex]. What are the coordinates of vertices [tex]$X' Y' Z'$[/tex]?

A.
\begin{array}{l}
X'(-6, -2) \\
Y'(2, 0) \\
Z'(-2, -6)
\end{array}

B.
\begin{array}{l}
X'(-6, 2) \\
Y'(2, 0) \\
Z'(-2, 6)
\end{array}

C.
\begin{array}{l}
X'(6, -2) \\
Y'(2, 0) \\
Z'(2, -6)
\end{array}

D.
\begin{array}{l}
X'(6, 2) \\
Y'(-2, 0) \\
Z'(2, 6)
\end{array}

Sagot :

GinnyAnswer
Let's solve the problem step-by-step:

1) Original coordinates:
- [tex]\( X(-6, -2) \)[/tex]
- [tex]\( Y(2, 0) \)[/tex]
- [tex]\( Z(-2, -6) \)[/tex]

2) Reflecting across the x-axis:
- To reflect a point [tex]\((a,b)\)[/tex] across the x-axis, you change the y-coordinate to its negative, which turns a point [tex]\((a, b)\)[/tex] into [tex]\((a, -b)\)[/tex].
- Reflecting [tex]\( X(-6, -2) \)[/tex] across the x-axis produces [tex]\( X'(-6, 2) \)[/tex].
- Reflecting [tex]\( Y(2, 0) \)[/tex] across the x-axis produces [tex]\( Y'(2, 0) \)[/tex].
- Reflecting [tex]\( Z(-2, -6) \)[/tex] across the x-axis produces [tex]\( Z'(-2, 6) \)[/tex].

3) Dilating by a scale factor of 2:
- To dilate a point [tex]\((a,b)\)[/tex] by a scale factor of 2 from the origin, you multiply both the x-coordinate and y-coordinate by 2, turning a point [tex]\((a, b)\)[/tex] into [tex]\((2a, 2b)\)[/tex].
- Dilating [tex]\( X'(-6, 2) \)[/tex] by a scale factor of 2 produces [tex]\( X''(-12, 4) \)[/tex].
- Dilating [tex]\( Y'(2, 0) \)[/tex] by a scale factor of 2 produces [tex]\( Y''(4, 0) \)[/tex].
- Dilating [tex]\( Z'(-2, 6) \)[/tex] by a scale factor of 2 produces [tex]\( Z''(-4, 12) \)[/tex].

Therefore, the coordinates of the vertices after reflection across the x-axis and dilation by a scale factor of 2 are:
- [tex]\( X''(-12, 4) \)[/tex]
- [tex]\( Y''(4, 0) \)[/tex]
- [tex]\( Z''(-4, 12) \)[/tex]

Hence, the results are [tex]\( X'' = (-12, 4), Y'' = (4, 0), Z'' = (-4, 12) \)[/tex].
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