![]() So we are done.(x,y)\rightarrow (−y,−x)\). Went from this point and now our blue is over the Sometimes we just want to write down the translation, without showing it on a graph. Made the reflection happen and notice it completely And then the other one is one, comma, one. So one is negative one,Ĭomma, negative five. The other is one, comma, one so let me see if I can remember that. We found two points thatĭefine that line of reflection so now let's use the tool to type them in. Is kind of the surface of the water, if you're That these are mirror images if this is kind of the mirror here. Side of the line equal distant and you get to its corresponding point on the image. This is the shortestĭistance from the line. Pick an arbitrary point on segment ME, say that point, and if you reflected over this line. So let's just first reflect point let me move this a little bit out of the way. And actually, let me just move this whole thing down here so that we can so that we can see what is going on a little bit clearer. We want to find the reflection across the X axis. ![]() Is a line of reflection 'cause you see that you So we can see the entire coordinate axis. I missed that magenta point a little bit, so let me go through the magenta point. Going to look something like, I want to draw this a littleīit straighter than that, it's going to look something like this. Let me draw the line of reflection, just 'cause we did all of this work, the line of reflection is Is a midpoint between these two so one, comma, one, just like that. Three minus one, three plus negative one, that's positive two over two is one. ![]() Let's see, negative five of plus seven is positive two, over two is one. Midpoint is going to be the average of the y coordinates. So the midpoint, the xĬoordinate of the midpoint, is going to be theĪverage of the x's here. The coordinates hereĪre x is equal to seven and y is equal to negative one. The coordinates of MĪre x is negative five, and y is equal to three. So now let's find the midpointīetween M and this point right over here. It looks like it'sĮquidistant between and E and this point right over here. So it's this point right over here and it does indeed look like the midpoint. So x is negative one, y is negative five. So this is going to be, this point right over here is going to be negative one, comma, negative five. For convenience, we shall rewrite x y 1 as y x 1. So we shall aim to show that algebraically. That's going to be the midpoint between E and the corresponding point on its image. Intuitively we know that the reflection would be the perpendicular of the line x 1. Let me do that in a blue color so you see where it came from. Negative four minus six which is going to be negative 10. We get the reflection (image) P ( x, y ) of a point P ( x, y) in the line AB: y m x c by demanding PP to be perpendicular AB and the foot of perpendicular N such that P N P N P P / 2 : x x m y y 1 2 m x y c 1 m 2. Negative four plus negative six, that's the same thing as Plus two is negative two, divided by two is negative one. Negative four plus negative six, over two and then close the parentheses. The y's, it's going to be negative four plus negative six over two. 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. The average of the x's and take the average of the y's. So what's the midpoint between negative four, negative four, and two, comma, negative six? Well you just have to take Two, x is equal to two, and y is equal to negative six. ![]() Right over here, that is, let's see that is x equals negative four, y is equal to negative four, and the coordinates for theĬorresponding point to E in the image. Midpoint between these two deep navy blue points. It, that line of reflection should contain the midpointīetween these two magenta points and it should contain the ![]() This and this should be E,Īnd this point should be equidistant from the line of reflection. This and this shouldīe equidistant from the line of reflection. And so between any pointĪnd its corresponding point on the image after the reflection, these should be equidistantįrom the line of reflection. We want to map point M to this point over here. We want to map point E, to this point right over here. One way to think about it, we want to map point E, To write something down so let me get my scratch pad out and I copied and pasted the same diagram. The line that we're going to reflect over with So if I click on this it says Reflection over the line from, and then we have two coordinate pairs. Let's see what they expect from us if we want to add a Reflection. This segment over here and we want to use a Reflection. Will map line segment ME, line segment ME, onto the Use the "Reflection" tool to define a reflection that ![]()
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