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aroub
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?
perpendicular bisector of BC
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? :)
Sure. You are welcome.
do u know how to solve this algebraically ?
how to get to 'perpendicular bisector of BC'
Emmm, i don't think so.. like what do you mean? show me? :)
|dw:1352049933632:dw|
Yea..the point A can be above BC or below BC according to that figure
ok, i choose point B as origin of my co-ordinate axes and point C on x-axis. as they are fixed i can put them wherever i want. now let co-ordinates 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, \(\huge x^2+y^2=(x-a)^2+y^2\) can u simplify this and tell me what u get ?
|dw:1352050205624:dw|
Oh, oh! No.. not yet at least. Let me know how to find the locus first :P Thank you though!!
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 x-axis) hence, 'perpendicular' did u get my explanation ?
A small correction @hartnn It is \[x ^{2}+y ^{2} = x ^{2} + (y-a)^{2}\]
Ok got it..that isn't the correction..the correction is B=(a,0)
correction is in diagram! not there |dw:1352050580220:dw|
Ya ya..that's what I was talking about
Yes I did! So you can get it with solving.. Not by memorizing the conditions or cases. Nice!