Find the image of [tex][tex]$(0,0)$[/tex][/tex] after two reflections, first across line 1 and then across line 2:

Line 1: [tex]y=3[/tex]
Line 2: [tex]x[/tex]-axis

A. [tex](0,6)[/tex]
B. [tex](0,-6)[/tex]
C. [tex](-6,0)[/tex]
D. [tex](6,0)[/tex]



Answer :

Let's find the image of the point [tex]\((0, 0)\)[/tex] after two reflections, first across the line [tex]\(y = 3\)[/tex] and then across the [tex]\(x\)[/tex]-axis.

1. First Reflection across the line [tex]\( y = 3 \)[/tex]:

When reflecting a point [tex]\((x, y)\)[/tex] across a horizontal line [tex]\( y = k \)[/tex], the rule is: [tex]\[ (x, y) \rightarrow (x, 2k - y) \][/tex]
For our specific case:
- The point is [tex]\((0, 0)\)[/tex].
- The line of reflection is [tex]\( y = 3 \)[/tex].
- Using the reflection rule: [tex]\[ (0, 0) \rightarrow (0, 2(3) - 0) = (0, 6) \][/tex]

So, after the first reflection, the point [tex]\((0, 0)\)[/tex] transforms to [tex]\((0, 6)\)[/tex].

2. Second Reflection across the [tex]\( x \)[/tex]-axis:

Reflecting a point [tex]\((x, y)\)[/tex] across the [tex]\( x \)[/tex]-axis involves the rule: [tex]\[ (x, y) \rightarrow (x, -y) \][/tex]
Applying this rule to our new point [tex]\((0, 6)\)[/tex]:
- The point is [tex]\((0, 6)\)[/tex].
- Reflecting across the [tex]\( x \)[/tex]-axis: [tex]\[ (0, 6) \rightarrow (0, -6) \][/tex]

So, after the second reflection, the point [tex]\((0, 6)\)[/tex] transforms to [tex]\((0, -6)\)[/tex].

Hence, the image of the point [tex]\((0, 0)\)[/tex] after the two reflections is [tex]\((0, -6)\)[/tex].

Therefore, the final result is [tex]\((0, 6)\)[/tex] after the first reflection and [tex]\((0, -6)\)[/tex] after the second reflection.

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