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aroub
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Given a triangle ABC such that B and C are fixed. Point A varies such that segment AB is congruent to segment AC. Find the locus of A.
Umm, an angle bisector maybe?
 2 years ago
 2 years ago
aroub Group Title
Given a triangle ABC such that B and C are fixed. Point A varies such that segment AB is congruent to segment AC. Find the locus of A. Umm, an angle bisector maybe?
 2 years ago
 2 years ago

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AbhimanyuPudi Group TitleBest ResponseYou've already chosen the best response.2
perpendicular bisector of BC
 2 years ago

aroub Group TitleBest ResponseYou've already chosen the best response.0
Aoh . I get it now! Because they said that the sides are congruent so they are equidistant so it's perpendicular bisector!! Let me draw it and tell me if it's right or wrong, okay? :)
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
Sure. You are welcome.
 2 years ago

aroub Group TitleBest ResponseYou've already chosen the best response.0
dw:1352049851819:dw
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
do u know how to solve this algebraically ?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
how to get to 'perpendicular bisector of BC'
 2 years ago

aroub Group TitleBest ResponseYou've already chosen the best response.0
Emmm, i don't think so.. like what do you mean? show me? :)
 2 years ago

AbhimanyuPudi Group TitleBest ResponseYou've already chosen the best response.2
dw:1352049933632:dw
 2 years ago

aroub Group TitleBest ResponseYou've already chosen the best response.0
Does it matter?
 2 years ago

AbhimanyuPudi Group TitleBest ResponseYou've already chosen the best response.2
Yea..the point A can be above BC or below BC according to that figure
 2 years ago

aroub Group TitleBest ResponseYou've already chosen the best response.0
Okay, thank youu!
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
ok, i choose point B as origin of my coordinate axes and point C on xaxis. as they are fixed i can put them wherever i want. now let coordinates of A be (x,y) we need a locus in terms of x and y let me draw it. dw:1352050085145:dw now AB=AC will give you, using distance formula, \(\huge x^2+y^2=(xa)^2+y^2\) can u simplify this and tell me what u get ?
 2 years ago

AbhimanyuPudi Group TitleBest ResponseYou've already chosen the best response.2
welcome :)
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
dw:1352050205624:dw
 2 years ago

aroub Group TitleBest ResponseYou've already chosen the best response.0
Oh, oh! No.. not yet at least. Let me know how to find the locus first :P Thank you though!!
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
if u simplify that u get x=a/2 which is a line that bisects BC , hence 'bisector' and is perpendicular to BC (as it is vertical line, its perpendicular to xaxis) hence, 'perpendicular' did u get my explanation ?
 2 years ago

AbhimanyuPudi Group TitleBest ResponseYou've already chosen the best response.2
A small correction @hartnn It is \[x ^{2}+y ^{2} = x ^{2} + (ya)^{2}\]
 2 years ago

AbhimanyuPudi Group TitleBest ResponseYou've already chosen the best response.2
Ok got it..that isn't the correction..the correction is B=(a,0)
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
correction is in diagram! not there dw:1352050580220:dw
 2 years ago

AbhimanyuPudi Group TitleBest ResponseYou've already chosen the best response.2
Ya ya..that's what I was talking about
 2 years ago

aroub Group TitleBest ResponseYou've already chosen the best response.0
Yes I did! So you can get it with solving.. Not by memorizing the conditions or cases. Nice!
 2 years ago
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