Isn't it easier to use the M co-ord's and drop vertical and horizontal construction line onto the X & Y axis say C & D and then demo that they are MP's of OA & OB which makes triangles OMA and OMB both equilateral and by definition OM = AM?
Is this valid? It seemed the obvious choice to me.
oh there are definitely other/easier ways to have done that then the way i did it. I just wanted to prove it using the midpoint and distance formulas.
as for your proof, if i understood you correctly, I don't think proving that C and D are the midpoints of OA and OB proves/implies that triangles OMA and OMB are equilateral or that OM=AM.
I guess what I didn't make clear was that the construction lines from the MP of AB to points C on the Y axis and D on the X axis will be at right angles to both axis. So the triangles OCM & ACM will be RA triangles with a common side CM and equal length bases AC & CO. Therefore their hyp. will be equal. It is in effect the same proof. I guess it is geometric rather than trigonometric.
seems to me like o,a,m was a perfect triangle, couldnt you just see if the angles were all the same. and if its a perfect triangle then the lines have to be perfect.
See, I don't even need a book. All I need is videos like this, and I'll be able to get through school with A's and B's like nerds do.
MejJalTok 11 months ago
G'day Chycho,
great series.
Isn't it easier to use the M co-ord's and drop vertical and horizontal construction line onto the X & Y axis say C & D and then demo that they are MP's of OA & OB which makes triangles OMA and OMB both equilateral and by definition OM = AM?
Is this valid? It seemed the obvious choice to me.
Keep Em Coming - Cheers - Mark
TheRealJavahead 2 years ago
oh there are definitely other/easier ways to have done that then the way i did it. I just wanted to prove it using the midpoint and distance formulas.
as for your proof, if i understood you correctly, I don't think proving that C and D are the midpoints of OA and OB proves/implies that triangles OMA and OMB are equilateral or that OM=AM.
chychochycho 2 years ago
I guess what I didn't make clear was that the construction lines from the MP of AB to points C on the Y axis and D on the X axis will be at right angles to both axis. So the triangles OCM & ACM will be RA triangles with a common side CM and equal length bases AC & CO. Therefore their hyp. will be equal. It is in effect the same proof. I guess it is geometric rather than trigonometric.
TheRealJavahead 2 years ago
this was a good lesson. i love math so much!
shennille9utube 3 years ago
nice lesson
packers489 3 years ago
Hello.
I live in the UK and you do the same sort of coordinate geometry we do.
You are a good teacher.
jrowlands90 4 years ago
??? m is mid point of a,b which makes it part of that object. not a refrence at all.
23remi23 4 years ago
seems to me like o,a,m was a perfect triangle, couldnt you just see if the angles were all the same. and if its a perfect triangle then the lines have to be perfect.
23remi23 4 years ago
then it would mean that you are assuming that the drawing is to scale ... you can never assume that ... the drawing is just used for reference
chychochycho 4 years ago
Thanks for the lesson!! You are a great teacher
afrothundah87 4 years ago