Review the basics of reflections, and then perform some reflections.

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Ultimate Hope

6 years agoPosted 6 years ago. Direct link to Ultimate Hope's post “Hw do I make the line go ...”

Hw do I make the line go where I want it, I'M SO CONFUSED!?

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(39 votes)

StellaAndShardBear

4 years agoPosted 4 years ago. Direct link to StellaAndShardBear's post “To move the line where yo...”

To move the line where you want it to be, click/tap and hold down the dot to move it. Hope this helps!

(7 votes)

Valerie

7 years agoPosted 7 years ago. Direct link to Valerie's post “a little bit troubling so...”

a little bit troubling some tips plz

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(33 votes)

A B

6 years agoPosted 6 years ago. Direct link to A B's post “How to do the practice: ...”

How to do the practice:

For the question: "Use the "Reflect" tool to find the image of MN for a reflection over the line y=-x+1".

First hit the "Reflect" button.

See AlsoReflecting shapes (article) | Reflections | Khan AcademyReflections in math. Formula, Examples, Practice and Interactive Applet on common types of reflections like x-axis, y-axis and lines:8.14: Rules for ReflectionsTriangle Reflection - Definition, Techniques, and Examples - The Story of Mathematics - A History of Mathematical Thought from Ancient Times to the Modern DayThen move the line to the center of the grid by moving your mouse over the line, away from the arrows, until the cursor changes to the icon you recognize as allowing for moving. Move the line to the center of the grid. Then decide the slope the line needs to have. Here it is -1, so use the rotation arrows to change the slope to -1. Then determine the y-intercept. Here that is 1. So, move the line so that it goes through the y-intercept (0,1). Finally, hit one of the arrows on either side of the line to reflect the image.

(5 votes)

Elena Kolesneva

6 years agoPosted 6 years ago. Direct link to Elena Kolesneva's post “i dont understand the lin...”

i dont understand the line of reflection in a form of an equation. there's smth missing here. is there a video?

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(15 votes)

JAYDEN JONES

2 years agoPosted 2 years ago. Direct link to JAYDEN JONES's post “understand that the same ...”

understand that the same distance away from the x-axis and the y-axis. there you will find your answer. keep practicing.

(1 vote)

Darren Drake

5 years agoPosted 5 years ago. Direct link to Darren Drake's post “Hi There.In the "Perfor...”

Hi There.

In the "Performing Reflections" I see the conventional equation is y=mx +b

Then the first example below it gives: y=x

which means that "the y intercept is 0 and the slope is 1".Is there a video explaining how the slope is determined for the line of reflection? It feels like a formula, or equation, with rules that make no sense to me. <grin>

I can reflect shapes, and rotate shapes, across a line of reflection - no problem. I cannot see how the line of reflection is originally determined from the formula. Particularly, the slope.

What am I missing? I might have jumped a step somewhere, I don't know.

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(9 votes)

kubleeka

5 years agoPosted 5 years ago. Direct link to kubleeka's post “Take a point A, and refle...”

Take a point A, and reflect it across a line so that it lands at B. Join segment AB. The reflecting line will be a perpendicular bisector of AB.

So if you know the coordinates of A and B, you can determine the slope of AB. Because they're perpendicular, you can then determine the slope of the reflecting line.

Also, since you know the coordinates of A and B, you can find their midpoint, which will be on the reflecting line. So now you have the slope of the reflecting line and a point on it, and can find an equation for it.

allecoth3546

2 years agoPosted 2 years ago. Direct link to allecoth3546's post “how did everyone find thi...”

how did everyone find this so hard! It's so easy.

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(9 votes)

cooperbadeer

4 years agoPosted 4 years ago. Direct link to cooperbadeer's post “im never gonna need to kn...”

im never gonna need to know this

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(0 votes)

David Severin

4 years agoPosted 4 years ago. Direct link to David Severin's post “You will need it to pass ...”

You will need it to pass high school and maybe college. Artists, Architects, Designers (including video game designers), and others need to know it also.

(14 votes)

Alvin Izera

4 years agoPosted 4 years ago. Direct link to Alvin Izera's post “what if a value of y is g...”

what if a value of y is given like....reflect across y=2

then ?? how to solve this?•

(4 votes)

Ian Pulizzotto

4 years agoPosted 4 years ago. Direct link to Ian Pulizzotto's post “Good question!If we ref...”

Good question!

If we reflect about the line y = 2, then the original point and its image have the same x-coordinate and have y-coordinates that average to 2 (and so add to twice 2, or 4).

So the image of any point (x, y) would be (x, 4-y). For example, the image of (6, 5) would be (6, 4-5) = (6, -1).

More generally, the image of any point (x, y) under reflection about the line y=b would be (x, 2b-y). Similarly, the image of any point (x, y) under reflection about the line x=a would be (2a-x, y). The concept of averaging in one coordinate and equality in the other coordinate leads to these formulas.

(5 votes)

harundiyarip

4 years agoPosted 4 years ago. Direct link to harundiyarip's post “your videos makes me smar...”

your videos makes me smarter, THANK YOU i appreciate it

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(6 votes)

christopher.shinn

7 years agoPosted 7 years ago. Direct link to christopher.shinn's post “i had some trouble with t...”

i had some trouble with these

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(5 votes)

28040

10 months agoPosted 10 months ago. Direct link to 28040's post “what job is this for?”

what job is this for?

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(5 votes)

660832

2 months agoPosted 2 months ago. Direct link to 660832's post “what does it mean if you...”

what does it mean if you did the problem correct but it says it is wrong ?

(1 vote)