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## 8th grade

# Reflections review

CCSS.Math:

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.*

## Want to join the conversation?

- Hw do I make the line go where I want it, I'M SO CONFUSED!?(34 votes)
- To move the line where you want it to be, click/tap and hold down the dot to move it. Hope this helps!(5 votes)

- a little bit troubling some tips plz(31 votes)
- 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)

- i dont understand the line of reflection in a form of an equation. there's smth missing here. is there a video?(15 votes)
- understand that the same distance away from the x-axis and the y-axis. there you will find your answer. keep practicing.(2 votes)

- 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)- 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.(4 votes)

- what if a value of y is given like....reflect across y=2

then ?? how to solve this?(4 votes)- 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)

- i had some trouble with these(5 votes)
- your videos makes me smarter, THANK YOU i appreciate it(5 votes)
- How do I reflect it if the reflection line is not directly through the diagonals?(5 votes)
- This is really easy is you know what to do(4 votes)
- i kind of understand it but im still having a hard time(2 votes)
- Could you please explain your problem? We might be able to help you then.(3 votes)