Main content

### Course: High school geometry > Unit 1

Lesson 5: Reflections# Determining reflections (advanced)

Sal is given two line segments on the coordinate plane, and determines the reflection that maps one of them into the other.

## Want to join the conversation?

- I have a test in a few weeks and I like to check my work over and over again, So, is there any other way to do it so I can check if the answers from all the strategies to see if they match?(23 votes)
- Not that I know of, but like Sal did at4:45, you can draw the line across and measure the units on either side of the line of reflection. They should come out equally if you are correct. Just be sure to draw the line correctly though!(13 votes)

- How does this guy have such good handwriting?(17 votes)
- Every time he says similarly incorrectly he writes a random number from 1 - 10(9 votes)

- Is there a faster way to find a point (-5, 1) with a line of reflection of y = -3 without graphing it?(5 votes)
- Yes there is! All you have to do is try to imagine the points in your head and the line you're reflecting on. Also make sure you have good knowledge of each quadrant on the coordinate plane.

•remember quadrant 1 = -,+.

•quadrant 2 = +,+

•quadrant 3 = -,-

•quadrant 4 = +,-

Oh and sorry I'm 7 years late(7 votes)

- Who knew numbers were vain enough to use mirrors...(8 votes)
- How would i reflect a point if it told me to reflect about the given line of y=1/2x + 16(7 votes)
- To reflect a point across a given line, you can use the following steps:

Find the slope of the given line.

Determine the negative reciprocal of the slope. This will be the slope of the perpendicular line.

Use the perpendicular slope and the given line's equation to find the equation of the perpendicular bisector. This bisector is the line about which the reflection occurs.

Find the intersection point of the given line and the perpendicular bisector. This point is the center of reflection.

Use the center of reflection to calculate the distance between the given point and the center.

Reflect the point across the center using the distance and direction.

Let's go through an example:

Suppose you have a point P(x, y) and the line

�

=

1

2

�

+

16

y=

2

1

x+16 is the line of reflection.

The slope of the given line is

1

/

2

1/2.

The negative reciprocal of

1

/

2

1/2 is

−

2

−2, so the slope of the perpendicular bisector is

−

2

−2.

Use the point-slope form to find the equation of the perpendicular bisector. Let's say the point P is

(

�

0

,

�

0

)

(x

0

,y

0

):

�

−

�

0

=

−

2

(

�

−

�

0

)

y−y

0

=−2(x−x

0

)

Find the intersection point of

�

=

1

2

�

+

16

y=

2

1

x+16 and the perpendicular bisector. This point is the center of reflection.

Calculate the distance between the given point P and the center of reflection.

Reflect the point across the center using the distance and direction.(1 vote)

- What does it mean when a line intersects?(1 vote)
- When a line intersects, two lines cross each other. An example is a perpendicular line. Two lines intersect creating a 90 degree angle.(8 votes)

- At2:20how did you know that it was over 2? Like how did you get that?(2 votes)
- Hi Hailey!

Sal is taking the average of the x coordinates and the average of the y coordinates. To calculate a non-weighted average, you add the appropriate numbers together and divide by how many different numbers you have. So if I wanted the average of the set {2, 3, 4, 5} you would do it this way:

2 + 3 + 4 +5 = 14

There are 4 addends (numbers that are being added together) here, so we will divide 14 by 4.

14/4 = 3.5

Therefore, the average {2, 3, 4, 5} is 3.5.

Here the 2 x coordinates are -4 and 2, and the y coordinates are -4 and -6.

Sal adds -4 + 2 = -2, -2 / 2 = -1.

-4 + -6 = -10, -10 / 2 = -5.

These are the coordinates of one endpoint of the line of reflection.

I hope this helped you!

Have a great day!(10 votes)

- Couldn't you just use triangles to find the midpoints?(4 votes)
- Yes, you can use triangles to find midpoints. In geometry, the midpoint of a line segment can be found by creating two triangles with the given line segment. The midpoint is the point where the two medians of the triangles intersect. The midpoint coordinates can be calculated using the average of the corresponding coordinates of the endpoints of the line segment.(2 votes)

- At2:01, you found the average of the points. Yet when I did this on the " Determine Reflection " in the MAP recommended practice Geometry > 231, and used this method,it wouldn't take it for a answer. I even went through the hints and used those answers, it would not take it. Where do I report things like this?(3 votes)
- this is not free v-bucks but also im confused(4 votes)

## Video transcript

- [Voiceover] We're asked to
use the "Reflection" tool to define a reflection that
will map line segment ME, line segment ME, onto the
other line segment below. So we want to map ME to
this segment over here and we want to use a Reflection. Let's see what they expect from us if we want to add a Reflection. So if I click on this it says Reflection over the line from, and then we have two coordinate pairs. So they want us to define
the line that we're going to reflect over with
two points on that line. So let's see if we can do that. To do that I think I need
to write something down so let me get my scratch pad out and I copied and pasted the same diagram. And the line of reflection,
one way to think about it, we want to map point E,
we want to map point E, to this point right over here. We want to map point M to this point over here. And so between any point
and its corresponding point on the image after the reflection, these should be equidistant
from the line of reflection. This and this should
be equidistant from the line of reflection. This and this should be E,
and this point should be equidistant from the line of reflection. Or another way of thinking about
it, that line of reflection should contain the midpoint
between these two magenta points and it should contain the
midpoint between these two deep navy blue points. So let's just calculate the midpoints. So we could do that with a
little bit of mathematics. The coordinates for E
right over here, that is, let's see that is x equals negative four, y is equal to negative four, and the coordinates for the
corresponding point to E in the image. This is x is equal to
two, x is equal to two, and y is equal to negative six. So what's the midpoint between negative four, negative four, and two, comma, negative six? Well you just have to take
the average of the x's and take the average of the y's. Let me do that, actually
I'll do it over here. So if I take the average
of the x's it's going to be negative four, negative four, plus two, plus two, over two, that's the average of the x's. And then the average of
the y's, it's going to be negative four plus negative six over two. Negative four plus negative six, over two and then close the parentheses. Let's see, negative four
plus two is negative two, divided by two is negative one. So it's going to be negative one, comma. Negative four plus negative six, that's the same thing as
negative four minus six which is going to be negative 10. Divided by two is negative five. Let me do that in a blue color so you see where it came from. Is going to be negative five. So there you have it. That's going to be the midpoint between E and the corresponding point on its image. So let's see if I can plot that. So this is going to be, this point right over here is going to be negative one, comma, negative five. So x is negative one, y is negative five. So it's this point right over here and it does indeed look like the midpoint. It looks like it's
equidistant between and E and this point right over here. And so this should sit on
the line of reflection. So now let's find the midpoint
between M and this point right over here. The coordinates of M
are x is negative five, and y is equal to three. The coordinates here
are x is equal to seven and y is equal to negative one. So the midpoint, the x
coordinate of the midpoint, is going to be the
average of the x's here. So let's see it's going
to be negative five plus seven over two. And the y coordinate of the
midpoint is going to be the average of the y coordinates. So three plus negative one over two. Let's see, negative five of plus seven is positive two, over two is one. Three minus one, three plus negative one, that's positive two over two is one. So the point one, comma, one
is a midpoint between these two so one, comma, one, just like that. So the line of reflection
is going to contain these two points. And two points define a line. Let me draw the line of reflection, just 'cause we did all of this work, the line of reflection is
going to look something like, I want to draw this a little
bit straighter than that, it's going to look something like this. And this makes sense that
this is a line of reflection. I missed that magenta point a little bit, so let me go through the magenta point. Okay, there you go. This makes sense that this
is a line of reflection 'cause you see that you
pick an arbitrary point on segment ME, say that point, and if you reflected over this line. This is the shortest
distance from the line. You just go onto the other
side of the line equal distant and you get to its corresponding point on the image. So it makes a lot of sense
that these are mirror images if this is kind of the mirror here. And you can image that this
is kind of the surface of the water, if you're
looking at it at an angle. I don't know if that helps you or not. But anyway we found two points. We found two points that
define that line of reflection so now let's use the tool to type them in. One is negative one, negative five. The other is one, comma, one so let me see if I can remember that. I have a bad memory. So one is negative one,
comma, negative five. And then the other one is one, comma, one. And we see it worked. We see it worked. When I did that, it actually
made the reflection happen and notice it completely
went from this point and now our blue is over the
image that we wanted to get to. So we are done.