From tech troubles to travel tips, IDNLearn.com has answers to all your questions. Get thorough and trustworthy answers to your queries from our extensive network of knowledgeable professionals.
Sagot :
To determine which reflection of the point [tex]\((0, k)\)[/tex] will produce an image at the same coordinates [tex]\((0, k)\)[/tex], let's analyze the effect of reflecting the point across each given axis and line.
1. Reflection across the [tex]\( x \)[/tex]-axis:
- For a point [tex]\((x, y)\)[/tex], reflecting it across the [tex]\( x \)[/tex]-axis results in the point [tex]\((x, -y)\)[/tex].
- For our specific point [tex]\((0, k)\)[/tex], reflecting it across the [tex]\( x \)[/tex]-axis would yield the point [tex]\((0, -k)\)[/tex], which is not the same as the original point unless [tex]\( k = 0 \)[/tex]. Therefore, this option does not generally give us [tex]\((0, k)\)[/tex].
2. Reflection across the [tex]\( y \)[/tex]-axis:
- For a point [tex]\((x, y)\)[/tex], reflecting it across the [tex]\( y \)[/tex]-axis results in the point [tex]\((-x, y)\)[/tex].
- For our specific point [tex]\((0, k)\)[/tex], reflecting it across the [tex]\( y \)[/tex]-axis would yield the point [tex]\((0, k)\)[/tex], which is exactly the same as the original point. Thus, this reflection maintains the coordinates.
3. Reflection across the line [tex]\( y = x \)[/tex]:
- For a point [tex]\((x, y)\)[/tex], reflecting it across the line [tex]\( y = x \)[/tex] results in the point [tex]\((y, x)\)[/tex].
- For our specific point [tex]\((0, k)\)[/tex], reflecting it across the line [tex]\( y = x \)[/tex] would yield the point [tex]\((k, 0)\)[/tex], which is different from the original point unless [tex]\( k = 0 \)[/tex]. Hence, this option does not generally give us [tex]\((0, k)\)[/tex].
4. Reflection across the line [tex]\( y = -x \)[/tex]:
- For a point [tex]\((x, y)\)[/tex], reflecting it across the line [tex]\( y = -x \)[/tex] results in the point [tex]\((-y, -x)\)[/tex].
- For our specific point [tex]\((0, k)\)[/tex], reflecting it across the line [tex]\( y = -x \)[/tex] would yield the point [tex]\((-k, 0)\)[/tex], which is once again different from the original point unless [tex]\( k = 0 \)[/tex]. So, this option also does not generally give us [tex]\((0, k)\)[/tex].
Upon examining all the reflections, the only reflection that results in the point [tex]\((0, k)\)[/tex] remaining unchanged is the reflection across the [tex]\( y \)[/tex]-axis.
Therefore, the correct reflection is across the [tex]\( y \)[/tex]-axis, which corresponds to the second option.
1. Reflection across the [tex]\( x \)[/tex]-axis:
- For a point [tex]\((x, y)\)[/tex], reflecting it across the [tex]\( x \)[/tex]-axis results in the point [tex]\((x, -y)\)[/tex].
- For our specific point [tex]\((0, k)\)[/tex], reflecting it across the [tex]\( x \)[/tex]-axis would yield the point [tex]\((0, -k)\)[/tex], which is not the same as the original point unless [tex]\( k = 0 \)[/tex]. Therefore, this option does not generally give us [tex]\((0, k)\)[/tex].
2. Reflection across the [tex]\( y \)[/tex]-axis:
- For a point [tex]\((x, y)\)[/tex], reflecting it across the [tex]\( y \)[/tex]-axis results in the point [tex]\((-x, y)\)[/tex].
- For our specific point [tex]\((0, k)\)[/tex], reflecting it across the [tex]\( y \)[/tex]-axis would yield the point [tex]\((0, k)\)[/tex], which is exactly the same as the original point. Thus, this reflection maintains the coordinates.
3. Reflection across the line [tex]\( y = x \)[/tex]:
- For a point [tex]\((x, y)\)[/tex], reflecting it across the line [tex]\( y = x \)[/tex] results in the point [tex]\((y, x)\)[/tex].
- For our specific point [tex]\((0, k)\)[/tex], reflecting it across the line [tex]\( y = x \)[/tex] would yield the point [tex]\((k, 0)\)[/tex], which is different from the original point unless [tex]\( k = 0 \)[/tex]. Hence, this option does not generally give us [tex]\((0, k)\)[/tex].
4. Reflection across the line [tex]\( y = -x \)[/tex]:
- For a point [tex]\((x, y)\)[/tex], reflecting it across the line [tex]\( y = -x \)[/tex] results in the point [tex]\((-y, -x)\)[/tex].
- For our specific point [tex]\((0, k)\)[/tex], reflecting it across the line [tex]\( y = -x \)[/tex] would yield the point [tex]\((-k, 0)\)[/tex], which is once again different from the original point unless [tex]\( k = 0 \)[/tex]. So, this option also does not generally give us [tex]\((0, k)\)[/tex].
Upon examining all the reflections, the only reflection that results in the point [tex]\((0, k)\)[/tex] remaining unchanged is the reflection across the [tex]\( y \)[/tex]-axis.
Therefore, the correct reflection is across the [tex]\( y \)[/tex]-axis, which corresponds to the second option.
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com is your go-to source for accurate answers. Thanks for stopping by, and come back for more helpful information.