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Lesson 4: Reflections

# 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 ${A}^{\prime }$ of $A\left(-6,7\right)$ under a reflection over $y=4$.

#### Solution

Step 1: Extend a perpendicular line segment from $A$ to the reflection line and measure it.
Since the reflection line is perfectly horizontal, a line perpendicular to it would be perfectly vertical.
Step 2: Extend the line segment in the same direction and by the same measure.
Answer: ${A}^{\prime }$ is at $\left(-6,1\right)$.

#### Practice problem

Draw the image of $B\left(7,-4\right)$ under a reflection over $x=2$.

#### Challenge problem

What is the image of $\left(-25,-33\right)$ under a reflection over the line $y=0$?
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### Let's study an example of reflecting over a diagonal line

We are asked to find the image ${C}^{\prime }$ of $C\left(-2,9\right)$ under a reflection over $y=1-x$.

#### Solution

Step 1: Extend a perpendicular line segment from $C$ 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 $\mathit{\text{1}}$ and $\mathit{\text{-1}}$ are always perpendicular.
For convenience, let's measure the distance in "diagonals":
Step 2: Extend the line segment in the same direction and by the same measure.
Answer: ${C}^{\prime }$ is at $\left(-8,3\right)$.

#### Practice problem

Draw the image of $D\left(3,-5\right)$ under a reflection over $y=x+2$.

#### Challenge problem

What is the image of $\left(-12,12\right)$ under a reflection over the line $y=x$?
$\left($
$,$
$\right)$

## Part 2: Reflecting polygons

### Let's study an example problem

Consider rectangle $EFGH$ drawn below. Let's draw its image ${E}^{\prime }{F}^{\prime }{G}^{\prime }{H}^{\prime }$ under a reflection over the line $y=x-5$.

#### 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 $E$, $F$, and $H$ were on an opposite side of the reflection line as $G$. The same is true about their images, but now they switched sides!
Now we simply connect the vertices.

#### Problem 1

Draw the images of the line segments $\stackrel{―}{IJ}$ and $\stackrel{―}{KL}$ under a reflection over $y=-3$.

#### Problem 2

Draw the image of $\mathrm{△}MNO$ under a reflection over $y=-1-x$.

## 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)?
• 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).
• I do not understand any of this at all. Is there an easier way to learn/understand it?
• 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!
• 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
Y= x-5
• 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!
• 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
• wow that is bonkers
• Is there a formula for the reflections?
• count the spaces between the line you are reflecting over
• 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.
• 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.

Hope that helps.
• why cant there be a video on this i dont understand it but a video would help
• There is a part that says "I want to see Sal doing a similar question" which helped me since I was having trouble.
• Seriously, this math stumps me
• isn't there an algebraic formula for this ?