WEBVTT
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Consider points π΄ two, three and π΅ negative four, negative three.
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Find the coordinates of πΆ given that πΆ is on the ray π΅π΄ but not on the segment π΄π΅ and π΄πΆ equals two π΄π΅.
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So the first thing Iβve drawn is Iβve drawn the points π΄ and π΅, and then also what weβve drawn here is the line segment π΄π΅.
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But now what we need to do is think about where the point πΆ is.
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Well, weβre told that itβs on the ray π΅π΄ but not on the line segment π΄π΅.
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But what does the ray π΅π΄ mean?
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Well, if we know that πΆ is on the ray π΅π΄, this means that from π΅ going towards π΄, πΆ is gonna lie along this line somewhere.
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However, we know that itβs not on the segment π΄π΅, so we know that itβs not in our line segment π΄π΅.
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So that means itβs not where the blue area is, so we know that it must extend beyond π΄.
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So what Iβve drawn here is a sketch of the scenario, just looking at the idea of a line.
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So weβre told that π΄ to πΆ is equal to two π΄π΅.
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So therefore, we know that our line is gonna be in the ratio one to two.
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And thatβs if we have the points π΅, π΄, and πΆ in that order.
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Now, the way we can approach this is to look at our π₯-coordinates and then look at our π¦-coordinates.
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So we start by looking at our π₯-coordinates.
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So what weβve done is labeled our coordinates π₯ sub two, π¦ sub two and π₯ sub one, π¦ sub one.
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Iβve just done it in this way around because weβre going from π΅ to π΄.
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So to find the difference between our π₯-coordinates, what weβre gonna do is π₯ sub two minus π₯ sub one, which is gonna be equal two minus negative four.
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Well, if we subtract a negative, itβs the same as adding a positive, so we get two plus four which is equal to six.
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So we know that the difference between π΅ and π΄ in the π₯-coordinate is six units.
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So now if we consider our ratio, which is one to two, we can see that the difference between the π₯-coordinates between π΄ and πΆ is gonna be twice that of π΅ to π΄, so we make two multiplied by six.
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So therefore, itβs going to be 12.
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So therefore, if we want to find out π₯ which weβre calling the π₯-coordinate of πΆ, then this is going to be the π₯-coordinate of π΄ which is two plus 12 because thatβs the difference between them, which is going to be equal to 14.
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Okay, great, so now weβre gonna look at the π¦-coordinates.
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So now what we want to do is find the difference between the π¦-coordinates.
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Itβs gonna be π¦ sub two minus π¦ sub one, which is gonna be three minus negative three.
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Well, if you subtract a negative, itβs the same as adding a positive like we said before.
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So once again, we can see the difference this time between the π¦-coordinates is six.
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So therefore, using the same rationale as before, thatβs our ratio one to two, we can see that, therefore, the difference between the π¦-coordinates of π΄ and πΆ must be double that.
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So itβs gonna be 12.
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So therefore, the π¦-coordinate of πΆ is going to be three, which was the π¦-coordinate of π΄ plus 12 which is gonna be equal to 15.
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So therefore, we can say that the coordinates of point πΆ given that πΆ is on the ray π΅π΄ but not on the segment π΄ to π΅ and π΄πΆ is equal to two π΄π΅ are 14, 15.