If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:7:19

Reflecting shapes: diagonal line of reflection

CCSS.Math:

Video transcript

line segments I end this segment i n over here and T oh this is T oh here are reflected over the line y is equal to negative X minus 2 so this is the line that they're reflected about this dashed purple line and it is indeed y equals negative X minus 2 this right over here is in slope-intercept form the slope should be negative 1 and we see that the slope of this purple line is indeed negative 1 if X changes by a certain amount and Y changes by the negative of that if X changes by 1 Y changes by negative 1 to get back to that line if X changes by positive 2 y changes by negative 2 to get back to another point on that line and the y-intercept we see when X is equal to 0 Y should be negative 2 when X is equal to 0 Y is indeed negative 2 so we validated that now to say draw the image of this reflection using the interactive graph all right so we can move these lines around and we want to reflect these and I could try to eyeball it you know maybe it's something like this I don't know it doesn't seem exactly right that looks close to the reflection of I N and for tio I'd want to move this down here tio looks like it would be I don't know I'm i balling it this is close but I can't be close I want to get exact so let's I've copied and pasted the original problem on my scratch pad so we can find the exact points and so I don't I don't just have to estimate this so let's go to the scratch pad so exactly what we just saw and the main realization is is if we want to reflect a given point if we want to reflect a given point say point I right over here what we want to do is we want to drop a perpendicular we want to find a line that's perpendicular a line that has the point I on it and it's perpendicular to this line right over here remember this is this is the line let me do this in the purple color this is the line y is equal to negative X minus 2 its slope its slope is equal to negative 1 so I want to a line that goes through a point I that has a that is perpendicular perpendicular to this line and I want to drop it I want to drop it to I want to drop it to the line that I'm going to reflect on and then I want to go the same distance on to the other side to find to find the corresponding point in the image so how do I do that well if this line if this purple line is a slope of negative one a line that is a line that is perpendicular to it a line that is perpendicular to it so this thing that I'm drawing in purple right over here it's slope is going to be the negative reciprocal of this so the reciprocal of negative one is still just negative one one over negative one is still negative one but we want the negative of that so the slope here needs to be one and luckily that's how I drew it the slope here needs to be equal to one which is however much I change in the X direction I change in the Y direction we see that to go from this point to this point right over here we decrease Y by four and we decrease X by four now if we want to stay on this line to find the reflection we just do the same thing we could decrease X by four so we're go from negative 2 to negative six and decrease Y by four and we end up at this point right over here so we end up at the point this is x equals negative 6 y is equal to negative 4 so this is this point corresponds to this point right over there now let's do the same thing let's do the same thing for point N for point n we already know if we drop a perpendicular if this is perpendicular it's going to have a slope of 1 because this purple line has a slope of negative 1 the negative reciprocal of negative 1 is positive 1 and let's see to go from this point to this point of intersection we have to go down 1 and 1/2 we're going down 1 and 1/2 and we're going to the left 1 and 1/2 so we want to do that on the other side we want to stay on this perpendicular line so we want to go left 1 and 1/2 and down 1 and 1/2 and we get to this point right over here which is the point x equals 3y is equal to negative 8 and so we are now equidistant we're on this perpendicular line still but we're equidistant on the other side so the image of I N is going to go through negative 6 comma negative 4 and 3 comma negative 8 so let me let me draw that so let me see if I can remember negative 6 negative 4 3 comma negative 8 so I have a bad memory so negative six negative four and three comma negative eight and I was close when I estimate it but I wasn't exactly right so that's looking that's looking pretty good and then actually we can do the exact same thing with points T in point O and let me do that so point T at the point t to get from point T to the line in the shortest distance once again we drop a perpendicular this line is going to have a slope of 1 because it's perpendicular to the line that has a slope of negative 1 and so to get there we have to decrease our X by we have to decrease our X we're going from x equals 5 to x equals looks like 1/2 so X went down by 4 and 1/2 in the X direction and Y also needs to go down by 4 and a half so if we want to stay on that line let's decrease our X by 4 and a half so that's 1/2 1 2 3 4 and Y needs to go down by 4 and a half so that's 1/2 1 2 3 4 and we get to this point right over here which is the point x equals negative 4 y is equal to negative 7 negative 4 comma negative 7 let me so this should be at negative 4 y equals negative 7 and there's a couple of things you could do here you could just say this is too long this is our two units long not like too long in length somehow this is two units long so maybe this is two units long so this is feeling good pretty good let's just go through the exercise 4.0 as well so point o once again this is going to have if we drop a perpendicular it's going to have a slope of 1 so whatever our change in X between this point and this point we're gonna have the same change in Y and our change in X to go from 7 to 7 to one and a half our change in X is five and a half so maybe let me do it this way so the change in X here so a change in X is equal to negative 5 and 1/2 5.5 if you subtract 5.5 from 7 you get to 1.5 and our change in Y our change in Y is also negative 5.5 change in Y is negative you're not going to see that something too different negative 5.5 and so we need to stand on this line so we want to change by those same amounts on to the other side of that line so if we decrease our X by 5 and 1/2 so 1/2 1 2 3 4 5 we get there and Y by 5 and 1/2 hat 1/2 1 2 3 4 5 we get to the point x equals negative 4 y is equal to negative 9 so we get to the point x equals negative 4 y is equal to negative 9 and we're done we have figured out the reflection of these two of these two segments