![]() ![]() ![]() ![]() And if you had some other triangle that had the same dimensions as this one, and this one can have any dimensions 'cause we haven't defined a, b, and c. Saying that the x-coordinate is a, and so it's a comma zero. And so to setup this proof, I put an arbitrary triangle here, but I put one vertex at the origin. Intersect in one point, but three lines intersecting Two different lines with different slopes to Of this video is to prove that the three medians ofĪ triangle always intersect at one point, which is pretty interesting, because you would expect I know that this took a long time to read but thank you for reading thus far. To fully understand how they got there, let's first talk about the difference between a "normal average" and a "weighted average." It's mesmerizing because it seems like magic that they looked like they appeared out of nowhere. Thank you for reading this far, I hope I somehow enlightened you. Notice we have two of c/2 and one of 0 so we can rewrite the expression to:Īnd there we have it, we've derived the expressions that Sal did. This also explains why we can say 0/3 = 1/3*(0) because we have one zero.Īnd there we have it, we derived the expression that Sal made himself. In other words, we have two of (a+b)/2 and one of zero. ![]() You might be thinking why we divided by 3 and that's because (a+b)/2 weighs two times than 0. Notice that we can 0/3 is the same thing as 1/3*(0) so now our expression would be: Since we know that (a+b)/2 weighs two more than 0 then the weighted average of the x-coordinate would be:Ĭombine like, or same, terms (imagine that (a+b)/2 is a term then our expression would now be): That would mean that that point weighs 2 times than point (0, 0). What does that mean? As you can see from the video, the intersection point is near to the point ((a+b)/2, c/2). Why? That's because we made it so that the number 7 is more important than our number 5 hence the weighted average is closer to 7.Ĥ:03 on the video and let's try to breakdown (2/3*((a+b)/2) + 1/3(0), 2/3*(c/2) + 1/3*(0)).Īnd remember that Sal said that the intersection point of the medians have the ratio of 2:3. Notice that we divided by four and not two because remember we said that 7 is three times more than 5, and there's only one 5, hence, three plus one is four, which is why we divided them by 4.Īlso, notice that our weighted average is now 6.5, which is 0.5 more than our normal average. Now what's your "weighted average" going to be? It's going to be (7+7+7+5)/4 or (3*7 + 5*1)/4 which equal to 6.5. (Remember that we're weighing 7 three times more than 5) What is the weighted average? It's still your average but taking into account that now your number 7 is more important than your number 5. However, what if, for some reason, the number 7 weighs three times than 5? In other words, what if the ratio between them is 3:4 where 7 weighs more? This is where weighted average comes in. What do you do? Of course you do the sum of those two and divide them by two, in other words, (7+5)/2 = 6. What is the normal average? That's your typical way of doing the average, which is also used for finding the midpoint of a line segment.įor example, you have two numbers 7 and 5, and you want to find their average. Before we dive in as to why they were multiplied that way, let's first talk about the difference between a "normal average" and a "weighted average." ![]()
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