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The base of a triangle is fixed and the ratio of the lenghts of the the other two sides is constant . Find the locus of the vertex joining the other two sides.
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\[\sqrt{x^2+y^2}=k \sqrt{(x-a)^2+y^2}\]
looks perfect !!
square on both sides n try to simplify...
yeahh checked with my calc :) but have trouble turning that into circle form..
got it... gimme sometime to post the solution
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