$A (3 x_{1}, 3 y_{1}), B (3 x_{2}, 3 y_{2}), C (3 x_{3}, 3 y_{3})$ are vertices of a triangle with orthocentre H at $(x_{1} + x_{2} + x_{3}, y_{1} + y_{2} + y_{3})$ then the $∠ A B C$

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Solution

The correct option is B
Centroid $G = (\frac{3 x_{1} + 3 x_{2} + 3 x_{3}}{3}, \frac{3 y_{1} + 3 y_{2} + 3 y_{3}}{3}) = (x_{1} + x_{2} + x_{3}, y_{1} + y_{2} + y_{3}) = H$
$∴ ∠ A B C = \frac{π}{3}$

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A(3x1,3y1),B(3x2,3y2),C(3x3,3y3) are vertices of a triangle with orthocentre H at (x1+x2+x3,y1+y2+y3) then the ∠ABC

The correct option is B Centroid G=(3x1+3x2+3x33,3y1+3y2+3y33)=(x1+x2+x3,y1+y2+y3)=H ∴∠ABC=π3

$A (3 x_{1}, 3 y_{1}), B (3 x_{2}, 3 y_{2}), C (3 x_{3}, 3 y_{3})$ are vertices of a triangle with orthocentre H at $(x_{1} + x_{2} + x_{3}, y_{1} + y_{2} + y_{3})$ then the $∠ A B C$

The correct option is B
Centroid $G = (\frac{3 x_{1} + 3 x_{2} + 3 x_{3}}{3}, \frac{3 y_{1} + 3 y_{2} + 3 y_{3}}{3}) = (x_{1} + x_{2} + x_{3}, y_{1} + y_{2} + y_{3}) = H$
$∴ ∠ A B C = \frac{π}{3}$