Grade 11Analytical Geometry
Q 16
Q 16
1 Answer
Yash Jain
Hey...its very easy. See, we know that for a triangle with vertices (x1,y1), (x2,y2) and (x3,y3) the centroid (h,k) is given as h = (x1+x2+x3)/3 and k = (y1+y2+y3)/3.
Here we have all the three vertices as (acost,asint), (bsint.-bcost) and (1,0). So h is
h = (acos+bsint+1)/3 or 3h-1 = acost+bsint. Similarly, 3k = asint-bcost.
Just square and add both the equations and you will get your answer, in terms of (x,y), as
(3x-1)2+(3y)2=a2+b2
Here we have all the three vertices as (acost,asint), (bsint.-bcost) and (1,0). So h is
h = (acos+bsint+1)/3 or 3h-1 = acost+bsint. Similarly, 3k = asint-bcost.
Just square and add both the equations and you will get your answer, in terms of (x,y), as
(3x-1)2+(3y)2=a2+b2
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