# Reflecting shapes (article)

Reflections Over the X-Axis and Y-Axis Explained!
Reflections Over the X-Axis and Y-Axis Explained!

## High school geometry

Reflecting shapes

Learn how to find the image of a given reflection.

In this article we will find the images of different shapes under different reflections.

## The line of reflection

A reflection is a transformation that acts like a mirror: It swaps all pairs of points that are on exactly opposite sides of the line of reflection.

The line of reflection can be defined by an equation or by two points it passes through.

## Part 1: Reflecting points

### Let’s study an example of reflecting over a horizontal line

We are asked to find the image of under a reflection over .

A coordinate plane. The x- and y-axes both scale by one. A horizontal line passes through four on the y axis. A point A is at negative six, seven.

#### Solution

Step 1: Extend a perpendicular line segment from to the reflection line and measure it.

Since the reflection line is perfectly horizontal, a line perpendicular to it would be perfectly vertical.

A coordinate plane. The x- and y-axes both scale by one. A horizontal line passes through four on the y axis. A point A is at negative six, seven. A ray extends from the point A three units to the dashed line.

Step 2: Extend the line segment in the same direction and by the same measure.

A coordinate plane. The x- and y-axes both scale by one. A horizontal line passes through four on the y axis. A point A is at negative six, seven. A ray extends from the point A three units to the dashed line. The ray passes the dashed line three more units.

A coordinate plane. The x- and y-axes both scale by one. A horizontal line passes through four on the y axis. A point A is at negative six, seven. A ray extends from the point A three units to the dashed line. The ray passes the dashed line three more units pointing at the image of a point A prime at negative six, one.

#### Challenge problem

What is the image of under a reflection over the line ?

### Let’s study an example of reflecting over a diagonal line

We are asked to find the image of under a reflection over .

A coordinate plane. The x- and y-axes both scale by one. A dashed line slants down from left to right passing through zero, one and one, zero. A point C is at negative two, nine.

#### Solution

Step 1: Extend a perpendicular line segment from to the reflection line and measure it.

Since the reflection line passes exactly through the diagonals of the unit squares, a line perpendicular to it should pass through the other diagonal of the unit square. In other words, lines with slopes and are always perpendicular.

For convenience, let’s measure the distance in “diagonals”:

A coordinate plane. The x- and y-axes both scale by one. A dashed line slants down from left to right passing through zero, one and one, zero. A point C is at negative two, nine. A ray extends from point C three diagonals to the dashed line.

Step 2: Extend the line segment in the same direction and by the same measure.

A coordinate plane. The x- and y-axes both scale by one. A dashed line slants down from left to right passing through zero, one and one, zero. A point C is at negative two, nine. A ray extends from point C three diagonals to the dashed line. The ray passes the dashed line and extends another three diagonals.

A coordinate plane. The x- and y-axes both scale by one. A dashed line slants down from left to right passing through zero, one and one, zero. A point C is at negative two, nine. A ray extends from point C three diagonals to the dashed line. The ray passes the dashed line and extends another three diagonals pointing at the image point C prime at negative eight, three.

#### Challenge problem

What is the image of under a reflection over the line ?

## Part 2: Reflecting polygons

### Let’s study an example problem

Consider rectangle drawn below. Let’s draw its image under a reflection over the line .

A coordinate plane. The x- and y-axes both scale by one. A dashed line slants up from left to right passing through zero, negative five and five, zero. A rectangle with the points E F G and H. Point E is at negative three, three, Point F is at five, three, Point G is at five, negative three. Point H is at negative three, negative three.

#### Solution

When we reflect a polygon, all we need is to perform the reflection on all of the vertices (this is similar to how we translate or rotate polygons).

Here are the original vertices and their images. Notice that , , and were on an opposite side of the reflection line as . The same is true about their images, but now they switched sides!

A coordinate plane. The x- and y-axes both scale by one. A dashed line slants up from left to right passing through zero, negative five and five, zero. Point E is at negative three, three, Point F is at five, three, Point G is at five, negative three. Point H is at negative three, negative three. A ray extends from point E through the dashed line to point E prime at eight, negative eight. A ray extends from point F through the dashed line to point F prime at eight, zero. A ray extends from point G through the dashed line to point G prime at two, zero. A ray extends from point H through the dashed line to point H prime at two, negative eight.

Now we simply connect the vertices.

A coordinate plane. The x- and y-axes both scale by one. A dashed line slants up from left to right passing through zero, negative five and five, zero. Point E is at negative three, three, Point F is at five, three, Point G is at five, negative three. Point H is at negative three, negative three. A ray extends from point E through the dashed line to point E prime at eight, negative eight. A ray extends from point F through the dashed line to point F prime at eight, zero. A ray extends from point G through the dashed line to point G prime at two, zero. A ray extends from point H through the dashed line to point H prime at two, negative eight. E prime, F prime, G prime, and H prime form the image of a rectangle.

#### Problem 1

Draw the images of the line segments and under a reflection over .

#### Problem 2

Draw the image of under a reflection over .

## Want to join the conversation?

• I understand how to algebraically perform reflections if the line of reflection is y = 0, x = 0, y = x, or y = -x. How can I algebraically perform a reflection for ANY line of reflection (e.g. how could I reflect (2, 9) across y = 7x + 2 algebraically)?(20 votes)
• Great question!
Let A be the point to be reflected, let k be the line about which the point is reflected, let B represent the desired point (image), and let C represent the intersection of line k and line AB. Note that line AB must be perpendicular to line k, and C must be the midpoint of segment AB (from the definition of a reflection).
So we can first find the equation of the line through point A that is perpendicular to line k. Then we can algebraically find point C, which is the intersection of these two lines. Then, using the fact that C is the midpoint of segment AB, we can finally determine point B.
Example: suppose we want to reflect the point A(2,9) about the line k with equation y = 7x + 2. So we first find the equation of the line through (2,9) that is perpendicular to the line y = 7x + 2. Since the line y = 7x + 2 has slope 7, the desired line (that is, line AB) has slope -1/7 as well as passing through (2,9).
So the desired line has an equation of the form y = (-1/7)x + b. Substituting the point (2,9) gives
9 = (-1/7)(2) + b which gives b = 65/7. So the equation of this line is y = (-1/7)x + 65/7.
Now we need to find the intersection of the lines y = 7x + 2 and y = (-1/7)x + 65/7 by solving this system of equations.
Using the substitution method gives 7x + 2 = (-1/7)x + 65/7; (50/7)x = 51/7; x = 51/50.
Then y = 7(51/50) + 2 = 457/50.
So the intersection of the two lines is the point C(51/50, 457/50). Recall that A is the point (2,9).
Since C is the midpoint of AB, we have
B = C + (C – A) = (51/50 + 51/50 – 2, 457/50 + 457/50 – 9) = (1/25, 232/25).
So the image (that is, point B) is the point (1/25, 232/25).(54 votes)
• I do not understand any of this at all. Is there an easier way to learn/understand it?(29 votes)
• Simple reflections are a matter of looking at a line and a point, line, or polygon on one side of it. You want to figure out the distance between the two and take that point, line, or polygon and put it the distance away from the line, but on the opposite side. Also, the line from any point A to its image A’ is perpendicular to the line of reflection. Hope this helps!(6 votes)
• I could really use Sal making a video about this, what’s written on this doc is really confusing.
Sometimes they explain things that are pretty basic and other times more complicated things they’ll just assume that we know them even though we haven’t covered it/them yet.
For instance I don’t understand what they mean when referring to the reflection line
Y=1-x
Y=x+2
• The reflection line is the line that you are reflecting over. Y=mx+b is just the basic slope-intercept equation. If you don’t understand slope -intercept, I recommend watching the videos Khan provides in the algebra courses. Since geometry tends to be taught after algebra in some cases, I think it’s why they didn’t explain it more in depth. Hope this helps!(7 votes)
• Is there a formula for the reflections?(21 votes)
• count the spaces between the line you are reflecting over(5 votes)
• I cried over this lesson for over an hour, took a 2 day break,and then cried at the thought of it. I finally went back to it with a fresh head and realized I had over thought the whole things and it really wasn’t that deep lol(15 votes)
• wow that is bonkers(2 votes)
• Is there a more mathematical way of calculating the reflection as opposed to manually counting on a graph? Perhaps using point slope (y=mx+b) or maybe by setting up a function? It seems like there should be a way to do this without requiring the graph.(6 votes)
• Yep, just plug the coordinates for each point into the point-slope equation that is given to get the reflected points.
Eg. In the last example in the article, you have the points M, O and N. To use M as an example, it’s graphed at (-7,2). In the question, it tells you that it is reflected over a line of the equation y=-1-x
To find the reflected coordinates of M using the point-slope equation given, plug in the following and solve:
2=-1-x (Gives you the X coordinate)
y=-1-(-7) (Gives you the Y coordinate)
You can repeat this step for the other points (O and N) and then graph it.
• why cant there be a video on this i dont understand it but a video would help(11 votes)
• There is a part that says “I want to see Sal doing a similar question” which helped me since I was having trouble.(2 votes)
• Seriously, this math stumps me(10 votes)
• x and y aren’t lines, they are still variables. But, when I say y = x, I can substitute a number for y and get the corresponding x value. So, if I take y = 5, I get x = 5. This gives me an (x,y) pair, which I can plot on a graph. When I plot a good amount of these pairs, you can see that they can be joined to form a line passing through the origin.
Similarly, we have x = 0. I can substitute a y value and get an x value for it. Where is the y, you ask? Well, let’s re-write the equation as x + 0y = 0. Now, thing is, whatever value of y you substitute, x will always be 0. So, all your points will be of the form (0,y). Plot those points and you’ll see that the line joining them is coincident to the y axis. You can expand the same logic to y = 0
Comment on Venkata’s post “x and y aren’t lines, the…”(9 votes)
• isn’t there an algebraic formula for this ?(3 votes)
• When you reflect a point across the line y = x, the x-coordinate and y-coordinate change places. If you reflect over the line y = -x, the x-coordinate and y-coordinate change places and are negated (the signs are changed). the line y = x is the point (y, x). the line y = -x is the point (-y, -x).(10 votes)

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