I didn't understand this formula intuitively either--I wondered why some other method than averages should not be used. But this proof is rock-solid and I think I will just stick with this formula because I now know precisely why it works. Thank you for this informative video!
I have a question, I don't know if I'm making sense but anyway. In the video a while ago, you have set the names for points A and B, I was wondering that is it okay to name it the other way? like example, in the video you have named point A as ( x , y1 ), I understand since it falls under the 'm' and 'P1'. So is it okay to use the x first from 'P1' then the y from 'm'? 'cause in the video you used the midpoint's x to be the x for the A and 'P1's y to be the y for A so it will be ( x1 , y). Same as for the B. Thank you I hope you reply, but it's okay if not and sorry for my bad grammar hope you understand what I'm trying to say.
I think you have taken this too far. You are unnecessarily complicating this proof. In order to be a good Mathematician, you need to simplify the situation, only use the amount of complication that is necessary. This lengthy theorem is not needed. I will show you simple proof without using congruent triangle. Just because you know congruent triangle theorem doesn't mean that you have to apply it everywhere. But want to give you credit that your methodology is good(naming and style). So the proof goes like this: The point is midpoint, so we have x-x1=x2-x, so we get 2x=x1+x2 or x=(x1+x2 )/2 and same way for y. So see in this case we don't need to use your complex Congruent theorem
Fyi, the point is a midpoint for the line joining the p1 and p2. You still need to use the congruent triangle theorem to prove x-x1=x2-x. Same goes for the y.
@@bonelesscookie1690 Exactly, because it is a midpoint, then x-x1 = x2-x. You don't need to use congruent triangle theorem to prove that. Like tell me how you're going to use that theorem when you don't even know what that point is (you don't have its coordinates, which you're trying TO FIND.) If I walk a distance, I can also walk half of it, right? The point where I walked half the distance is the mid point, if I walk another half distance, I'll have walked the entire distance, I don't need to prove that. If I translate this mathematically it means that: if I have two points A(x1,y1) and B(x2,y2), a distance must exist (from our previous obvious logic) called M with coordinates M(a,b) this means AM = BM --> Using distance formula, (a-x1) = (x2-a), Solve for "a" and get a=(x1+x2)/2. The same is true for y.
![30 ชั่วโมงคนเดียว บนรถไฟชั้น 3 โซเวียต [Around The World Ep. 2]](http://i.ytimg.com/vi/MlE9boAVqDA/mqdefault.jpg)
very elegant
I didn't understand this formula intuitively either--I wondered why some other method than averages should not be used. But this proof is rock-solid and I think I will just stick with this formula because I now know precisely why it works. Thank you for this informative video!
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Great stuff maam!
Finally a correct proof! TH-cam has too many "proofs" that assume the formula as part of their "proof". So thank you for making this video
Actually, the formula shall be derived from geometry and their coordinate. Before derivation, the mid-point formula is unknown.
Thank you so much! Super helpful.
thank you so much it really helped me a lot
Thank you, you helped me clarify some misunderstandings :)
Thanks very much for you I was searching too much for this prove☺️☺️
Thanks a lot. I'm looking for this proof.
🙏🙏👍👍
Oh my the algebra works
I have a question, I don't know if I'm making sense but anyway. In the video a while ago, you have set the names for points A and B, I was wondering that is it okay to name it the other way? like example, in the video you have named point A as ( x , y1 ), I understand since it falls under the 'm' and 'P1'. So is it okay to use the x first from 'P1' then the y from 'm'? 'cause in the video you used the midpoint's x to be the x for the A and 'P1's y to be the y for A so it will be ( x1 , y). Same as for the B. Thank you I hope you reply, but it's okay if not and sorry for my bad grammar hope you understand what I'm trying to say.
Yes you can
Thanks
Will you try to speak in Hindi
I think you have taken this too far. You are unnecessarily complicating this proof. In order to be a good Mathematician, you need to simplify the situation, only use the amount of complication that is necessary. This lengthy theorem is not needed. I will show you simple proof without using congruent triangle. Just because you know congruent triangle theorem doesn't mean that you have to apply it everywhere. But want to give you credit that your methodology is good(naming and style). So the proof goes like this: The point is midpoint, so we have x-x1=x2-x, so we get 2x=x1+x2 or x=(x1+x2 )/2 and same way for y. So see in this case we don't need to use your complex Congruent theorem
How do you know x-x1=x2-x
@@lohithakshkondaka9848 it was on the vedio that 'P1A=MB'
Fyi, the point is a midpoint for the line joining the p1 and p2. You still need to use the congruent triangle theorem to prove x-x1=x2-x. Same goes for the y.
KINDLY Take a look at MODERN'S VEGA - ( TH-cam link) excellent for MATHEMATICS LEARNING !
@@bonelesscookie1690 Exactly, because it is a midpoint, then x-x1 = x2-x. You don't need to use congruent triangle theorem to prove that. Like tell me how you're going to use that theorem when you don't even know what that point is (you don't have its coordinates, which you're trying TO FIND.)
If I walk a distance, I can also walk half of it, right? The point where I walked half the distance is the mid point, if I walk another half distance, I'll have walked the entire distance, I don't need to prove that. If I translate this mathematically it means that:
if I have two points A(x1,y1) and B(x2,y2), a distance must exist (from our previous obvious logic) called M with coordinates M(a,b)
this means AM = BM --> Using distance formula, (a-x1) = (x2-a), Solve for "a" and get a=(x1+x2)/2. The same is true for y.
Than derivation your unnecessary talks are more