![]() 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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