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aroub
 3 years ago
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?
aroub
 3 years ago
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?

This Question is Closed

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

aroub
 3 years ago
Best ResponseYou've already chosen the best response.0Aoh . 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? :)

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.0Sure. You are welcome.

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

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

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

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

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

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0ok, 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 ?

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352050205624:dw

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

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0if 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 ?

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

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

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

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

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