So that's half, one, two, three, four, and we get to this point right over here, which is the point X equals negative four, Y is equal to negative seven. And Y needs to go down by four and a half. So if we want to stay on that line, let's decrease our X by four and a half. So X went down by four andĪ half in the X direction and Y also needs to goĭown by four and a half. We're going from XĮquals five to X equals, it looks like, half. Have to decrease our X by we have to decrease our X. This line is going to have a slope of one, because it's perpendicular to the line that has a slope of negative one. Line in the shortest distance, once again, we drop a perpendicular. So point T, to point T, To get from point T to the And actually, we canĭo the exact same thing with points T and point O. And I was close when I estimated, but I wasn't exactly right. So negative six, negative four,Īnd three, negative eight. Let me see if I can remember, negative six, negative four,Īnd three, negative eight. So the image of IN is going to go through negative six, negative four,Īnd three, negative eight. We're on this perpendicular line, still, but we're equidistant on the other side. Over here, which is the point X equals three, Y isĮqual to negative eight. So we want to go left one and a half, and down one and a half. We're going down one and a half, and we're going to the And let's see, to go from this point to this point of intersection, we have to go down one and a half. Going to have a slope of one because this purple line For point N, we already know, if we drop a perpendicular, and this is perpendicular, it's So this is, this point corresponds to this point right over there. So we end up at the point, this is X equals negative six, Two to negative six, and decrease Y by four, and we end up at this We could decrease X by four, so we'll go from negative Now, if we want to stay on this line to find the reflection, This point right over here, we decrease Y by four and we decrease X by four. The slope here needs to be equal to one which is however much IĬhange in the X direction I change in the Y direction. So the reciprocal of negative one is still just negative one. In purple right over here, its slope is going to be the So how do I do that? Well if this line, if this purple line has a slope of negative one, a line that is perpendicular to it a line that is perpendicular to it so this thing that I'm drawing That I'm going to reflect on, and then I'm going to go the same distance onto the other side to find to find the corresponding Goes through I, point I, that is perpendicular to this line, and I want to drop it, I want to drop it to, I want to drop it to the line Remember this is the line, let me do this is that purple color. That's perpendicular, or a line that has the point I on it, and it's perpendicular 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. And the main realization is, is that if we want to
Pasted the original problem on my scratch pad so weĬan find the exact points and so I don't just have to estimate this. TO looks like it would be, I don't know, I'm eyeballing it. That looks close to the reflection of IN, and for TO I'd want to And I could try to eyeball it, you know, maybe it's something like this. These lines around, and we want to reflect these.
Image of this reflection, using the interactive graph." Alright, so we can move When X is equal to zero, Y is indeed negative two. And the Y intercept? We see when X is equal to zero, Y should be negative two. If X changes by positive two, Y changes by negative two to get back to another point on that line. If X changes by one, YĬhanges by negative one to get back to that line. If X changes by a certain amount, Y changes by the negative of that. Slope of this purple line is indeed negative one. The slope should be negative one, and we see that the They're reflected about this dashed, purple line. Is segment IN over here, and TO, this is TO here, are reflected over the line Y is equal to negative X minus two.
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