Equation of the locus of the centroid of the triangle
kvnmurty:
vertices : are these : (a cos t. a sin t), (b sin t, -b cos t) , (1, 0)
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Vertices of ΔABC: A(a cos t, a sin t), B (b sin t, - b cos t), C (1, 0)
Centroid G: x = (a cos t + b sin t + 1) / 3 --- (1)
y = (a sin t - b cos t + 0) /3 --- (2)
From these a cos t + b sin t = 3 x - 1
-b cos t + a sin t = 3 y
Hence, ( 3 x -1)² + (3 y)² = a² + b²
This is the locus.
Centroid G: x = (a cos t + b sin t + 1) / 3 --- (1)
y = (a sin t - b cos t + 0) /3 --- (2)
From these a cos t + b sin t = 3 x - 1
-b cos t + a sin t = 3 y
Hence, ( 3 x -1)² + (3 y)² = a² + b²
This is the locus.
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