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Reflections review

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

What is a reflection?

A reflection is a type of transformation that takes each point in a figure and reflects it over a line.
This reflection maps triangle, A, B, C onto the blue triangle over the gold line of reflection.
The result is a new figure, called the image. The image is congruent to the original figure.
Want to learn more about different types of transformations? Check out this video.

Performing reflections

The line of reflection is usually given in the form y, equals, m, x, plus, b.
Each point in the starting figure is the same perpendicular distance from the line of reflection as its corresponding point in the image.
Example:
Reflect start overline, P, Q, end overline over the line y, equals, x.
First, we must find the line of reflection y, equals, x. The slope is 1 and the y intercept is 0.
When the points that make up start overline, P, Q, end overline are reflected over the line y, equals, x, they travel in a direction perpendicular to the line and appear the same distance from the line on the other side.
Note that in the case of reflection over the line y, equals, x, every point left parenthesis, a, comma, b, right parenthesis is reflected onto an image point left parenthesis, b, comma, a, right parenthesis.
Reflecting over the line y, equals, x maps start overline, P, Q, end overline onto the blue line below.
Want to learn more about performing reflections? Check out this video.

Practice

Problem 1
Plot the image of triangle triangle, A, B, C under a reflection across the y-axis.

Want to try more problems like this? Check out this exercise.

Want to join the conversation?

  • aqualine ultimate style avatar for user Valerie
    a little bit troubling some tips plz
    (30 votes)
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    • boggle green style avatar for user A B
      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.

      Then 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.
      (4 votes)
  • orange juice squid orange style avatar for user Elena Kolesneva
    i dont understand the line of reflection in a form of an equation. there's smth missing here. is there a video?
    (15 votes)
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  • aqualine sapling style avatar for user Darren Drake
    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.
    (6 votes)
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    • leaf green style avatar for user kubleeka
      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.
      (3 votes)
  • blobby green style avatar for user Alvin Izera
    what if a value of y is given like....reflect across y=2
    then ?? how to solve this?
    (4 votes)
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    • primosaur seed style avatar for user Ian Pulizzotto
      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.
      (4 votes)
  • duskpin seedling style avatar for user christopher.shinn
    i had some trouble with these
    (5 votes)
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  • blobby green style avatar for user harundiyarip
    your videos makes me smarter, THANK YOU i appreciate it
    (5 votes)
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  • duskpin seedling style avatar for user Hannah Mendoza
    How do I reflect it if the reflection line is not directly through the diagonals?
    (5 votes)
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  • blobby green style avatar for user BerlJ
    I have tried to get this countless times and I can't get the answer
    (4 votes)
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  • blobby green style avatar for user jmamea99
    This is really easy is you know what to do
    (4 votes)
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  • aqualine seed style avatar for user Ultimate Hope
    Hw do I make the line go where I want it, I'M SO CONFUSED!?
    (2 votes)
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