If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

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 ABC onto the blue triangle over the gold line of reflection.
An XY coordinate plane with a triangle and a reflected copy of it graphed. The horizontal x axis runs left to right from negative 10 to 10 in intervals of 1. The vertical y axis runs up and down from negative 10 to 10 in intervals of 1. The triangle is reflected across the dashed line that runs diagonally from quadrant III to quadrant I through the origin. The original triangle has 3 vertices. Point A is plotted at (negative 3, 5), point B is plotted at (5, 9), and point C is plotted at (2, 4). The vertices of the reflected triangle are (5, negative 3), (9, 5), and (4, 2).
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=mx+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 PQ over the line y=x.
First, we must find the line of reflection y=x. The slope is 1 and the y intercept is 0.
An XY coordinate plane with a line segment graphed. The horizontal x axis runs left to right from negative 10 to 10 in intervals of 1. The vertical y axis runs up and down from negative 10 to 10 in intervals of 1. There is a dashed line that runs diagonally from quadrant III to quadrant I through the origin. The line segment begins at the solid point labeled P, which is plotted at (negative 7, negative 3), and ends at the solid point labeled Q, which is plotted at (3, 1).
When the points that make up PQ are reflected over the line y=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=x, every point (a,b) is reflected onto an image point (b,a).
An XY coordinate plane with a line segment graphed. The horizontal x axis runs left to right from negative 10 to 10 in intervals of 1. The vertical y axis runs up and down from negative 10 to 10 in intervals of 1. There is a dashed line that runs diagonally from quadrant III to quadrant I through the origin. The line segment begins at the solid point labeled P, which is plotted at (negative 7, negative 3), and ends at the solid point labeled Q, which is plotted at (3, 1). A solid point is plotted at (negative 3, negative 7). There is a dashed line that begins at point P and ends at the point at (negative 3, negative 7). A solid point is plotted at (1,3). There is a dashed line that begins at point Q and ends at the point at (1, 3).
Reflecting over the line y=x maps PQ onto the blue line below.
An XY coordinate plane with a line segment and its reflected copy of it graphed. The horizontal x axis runs left to right from negative 10 to 10 in intervals of 1. The vertical y axis runs up and down from negative 10 to 10 in intervals of 1. There is a dashed line that runs diagonally from quadrant III to quadrant I through the origin. The original line segment begins at the solid point labeled P, which is plotted at (negative 7, negative 3), and ends at the solid point labeled Q, which is plotted at (3, 1). The reflected line segment begins at the solid point that is plotted at (negative 3, negative 7), and ends at the solid point that is plotted at (1, 3).
Want to learn more about performing reflections? Check out this video.

Practice

Problem 1
Plot the image of triangle ABC under a reflection across the y-axis.

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

Want to join the conversation?

  • aqualine seed style avatar for user Ultimate Hope
    Hw do I make the line go where I want it, I'M SO CONFUSED!?
    (38 votes)
    Default Khan Academy avatar avatar for user
  • aqualine ultimate style avatar for user Valerie
    a little bit troubling some tips plz
    (32 votes)
    Default Khan Academy avatar avatar for user
    • 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.
      (5 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)
    Default Khan Academy avatar avatar for user
  • 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.
    (8 votes)
    Default Khan Academy avatar avatar for user
    • 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.
      (7 votes)
  • aqualine ultimate style avatar for user allecoth3546
    how did everyone find this so hard! It's so easy.
    (8 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user cooperbadeer
    im never gonna need to know this
    (1 vote)
    Default Khan Academy avatar avatar for user
  • 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)
    Default Khan Academy avatar avatar for user
    • 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.
      (5 votes)
  • blobby green style avatar for user harundiyarip
    your videos makes me smarter, THANK YOU i appreciate it
    (6 votes)
    Default Khan Academy avatar avatar for user
  • duskpin seedling style avatar for user christopher.shinn
    i had some trouble with these
    (5 votes)
    Default Khan Academy avatar avatar for user
  • 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)
    Default Khan Academy avatar avatar for user