Determining reflections (video) | Khan Academy (2024)

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  • ramona.spencer

    5 years agoPosted 5 years ago. Direct link to ramona.spencer's post “are there any tricks or r...”

    are there any tricks or rules with rigid transformations?

    (20 votes)

    • njeevan

      5 years agoPosted 5 years ago. Direct link to njeevan's post “I can't think of any tric...”

      Determining reflections (video) | Khan Academy (4)

      I can't think of any tricks, but I do know a rule:
      A rigid transformation only occours if the 2nd image of the shape preserves distance between points, and preserves the angle measure of the lines.

      (19 votes)

  • Barilugbene261

    5 years agoPosted 5 years ago. Direct link to Barilugbene261's post “How do change figure acr...”

    How do change figure across the y-axis

    (5 votes)

    • Polina Vitić

      5 years agoPosted 5 years ago. Direct link to Polina Vitić's post “To "*reflect*" a figure a...”

      Determining reflections (video) | Khan Academy (8)

      To "reflect" a figure across the y-axis, you want to do two things. For each of the figure's points:
      - multiply the x-value by -1
      - keep the y-value the same

      For instance, Triangle ABC (in the video) has the following three points:
      A (2, 6)
      B (5, 7)
      C (4, 4)

      To reflect Triangle ABC across the y-axis, we need to take the negative of the x-value but leave the y-value alone, like this:

      A (-2, 6)
      B (-5, 7)
      C (-4, 4)

      * Please note that the process is a bit simpler than in the video because the line of reflection is the actual y-axis. If the line of reflection was something else (like x = -4), you would need to do more than just taking the negative of the x-value - the process would be similar to what Sal does in the video.

      Hope this helps!

      (16 votes)

  • Mohammad Zayd

    5 years agoPosted 5 years ago. Direct link to Mohammad Zayd's post “I have a question. To fin...”

    I have a question. To find the line of reflection for a triangle, could someone count all the spaces between the two same vertices and then divide them by two. Then add that quotient to a vertice. One example could be in the video. The distance between Triangle ABC's vertice of C and Triangle A'B'C''s vertice of C is six. So then divide six by two to get 3. Then add that 3 to Triangle A'B'C' vertice c's Y-coordinate to get 1. The line of reflection is on the Y-coordinate of 1. Sorry if this was a little confusing. It is difficult to type about Triangle A'B'C' and the different vertices. Sorry.

    (10 votes)

    • Ellie

      5 months agoPosted 5 months ago. Direct link to Ellie's post “Yes, you can do it that w...”

      Yes, you can do it that way, although you probably figured that out by now because it's been 4 years.

      (5 votes)

  • mohidafzal31

    5 years agoPosted 5 years ago. Direct link to mohidafzal31's post “I can't seem to find it a...”

    I can't seem to find it anywhere, but one of the questions in a worksheet given by my teacher, we are asked to:
    Reflect at "y = -x"
    Is there a video or exercise on this that I missed? if not then pls guide me

    (6 votes)

    • mohidafzal31

      5 years agoPosted 5 years ago. Direct link to mohidafzal31's post “*Nevermind, punching y = ...”

      *Nevermind, punching y = -x into desmos gave me the line of reflection!*

      (7 votes)

  • bhudson642

    5 years agoPosted 5 years ago. Direct link to bhudson642's post “Why is there nothing on d...”

    Why is there nothing on dilation in this playlist? It's the only type of transformation not covered,

    (5 votes)

    • payal

      5 years agoPosted 5 years ago. Direct link to payal's post “there is, just keep going...”

      there is, just keep going down, it's the third to last group in this playlist

      (8 votes)

  • Aryanna Cortez

    2 years agoPosted 2 years ago. Direct link to Aryanna Cortez's post “Do you know any tricks or...”

    Do you know any tricks or like an easier way to find reflections?

    (3 votes)

    • The Telepath

      2 years agoPosted 2 years ago. Direct link to The Telepath's post “I use a memorization tric...”

      I use a memorization trick. Let's say you are given the point (2, -7).
      To reflect across the x-axis, use the rule (x, -y). This will give you (2, 7).
      To reflect across the y-axis, use the rule (-x, y). This gives you (-2, -7).
      To reflect across the line y=x, use the rule (y, x). This gives you (-7, 2).
      To reflect across the line y=-x, use the rule (-y, -x). This gives you (7, -2).

      Just memorize these formulas and you'll be good. You don't have to graph a point to find its reflection point.

      Hope this helps :D

      (7 votes)

  • zaksab1

    a year agoPosted a year ago. Direct link to zaksab1's post “i didn't understand”

    i didn't understand

    (3 votes)

    • joshua

      a year agoPosted a year ago. Direct link to joshua's post “Please specify what you d...”

      Please specify what you didn't understand. To do reflection for a shape, simply reflect each point respectively, last connect it, forming the reflected shape.

      To know where do you place the reflected point, simply count how many unit(s) is there from that initial point to the line of reflection. Then place the point on the other side of the line of reflection with the same number of unit(s).

      (5 votes)

  • Anderson Adoral

    a year agoPosted a year ago. Direct link to Anderson Adoral's post “what if the line of refle...”

    what if the line of reflection os oblique? is there a general rule for the points?

    (3 votes)

    • Venkata

      a year agoPosted a year ago. Direct link to Venkata's post “One thing you could do is...”

      One thing you could do is this: Consider the point given and the line of reflection (which is oblique). Now, draw a line from the point till you intersect the line of reflection. After you intersect it, draw a line perpendicular to the line you just drew, but make sure that this line is equal in length to the first line. Where your second line stops is the reflection of the point.

      Observe that the idea here is to make a square with the point as one corner and the line of reflection as the diagonal.

      (5 votes)

  • Anna Maxwell

    4 years agoPosted 4 years ago. Direct link to Anna Maxwell's post “So was that reflection a ...”

    So was that reflection a reflection across the y-axis?

    (2 votes)

    • Odelia

      4 years agoPosted 4 years ago. Direct link to Odelia's post “No, It would be a reflect...”

      No, It would be a reflection across something on the x-axis.
      Hope that helps!

      (5 votes)

  • MartiW

    10 months agoPosted 10 months ago. Direct link to MartiW's post “the dang volume isn't wor...”

    the dang volume isn't working, so, I am confused on what's happening. Please help me with this!

    (2 votes)

    • Storm_0891

      10 months agoPosted 10 months ago. Direct link to Storm_0891's post “Sal was just attempting t...”

      Sal was just attempting to find the where the line of reflection is at. He found it, then checked his work with the points by counting how many units away they were from the line.

      (3 votes)

Video transcript

- [Instructor] We're asked todraw the line of reflection that reflects triangle ABC,so that's this blue triangle, onto triangle A prime B prime C prime, which is this redtriangle right over here. And they give us alittle line drawing tool in order to draw the line of reflection. So the way I'm gonna think about it is well, when I just eyeball it, it looks like I'm just flipped over some type of a horizontal line here. But let's see if we can actually construct a horizontal line whereit does actually look like the line of reflection. So let's see, C and C prime, how far apart are they from each other? So if we go one, two,three, four, five, six down. So they are six apart. So let's see if we just putthis three above C prime and three below C, let's seeif this horizontal line works as a line of reflection. So C, or C prime isdefinitely the reflection of C across this line. C is exactly three units above it, and C prime is exactlythree units below it. Let's see if it works for A and A prime. A is one, two, three,four, five units above it. A prime is one, two, three,four, five units below it. So that's looking good. Now let's just check out B. So B, we can see it's at they-coordinate here is seven. This line right over hereis y is equal to one. And so what we wouldhave here is, let's see, this looks like it's sixunits above this line, and B prime is six units below the line. So this indeed works. We've just constructedthe line of reflection that reflects the bluetriangle, triangle ABC, onto triangle A prime B prime C prime.

Determining reflections (video) | Khan Academy (2024)
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