Question

if theta is a parameter is equals to a cos theta cos a sin theta and b is equals to B sin theta minus B cos theta is equals to 10 then the locus of centroid of triangle ABC is ​

Answer 1

Answer:

Let (h,k) be the coordinates of the centroid of ∆ABC.

Thus 3h=cosa+sina+1

         3k=sina-cosa+2

=>(3h-1)^2+(3k-2)^2=2

Hence locus of (h,k) is

(3x-1)^2+(3y-2)^2=2

9x^2+9y^2-6x-12y+3=0

3x^2+3y^2-2x-4y+1=0

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