The orthocentre of a Delta ABC is B and the c | vedclass

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(A) In a triangle,the orthocentre $H$,centroid $G$,and circumcentre $S$ are collinear,and $G$ divides $HS$ in the ratio $2:1$. Given orthocentre $H =

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$(2a, 2b)$

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(A) In a triangle,the orthocentre $H$,centroid $G$,and circumcentre $S$ are collinear,and $G$ divides $HS$ in the ratio $2:1$. Given orthocentre $H = B$ and circumcentre $S = (a, b)$. Since $A$ is the origin $(0, 0)$,the centroid $G$ is $\frac{A+B+C}{3} = \frac{0+B+C}{3} = \frac{B+C}{3}$. Using the property $G = \frac{H+2S}{3}$,we have $\frac{B+C}{3} = \frac{B+2S}{3}$. This implies $B+C = B+2S$,so $C = 2S$. Given $S = (a, b)$,the coordinates of $C$ are $(2a, 2b)$.

The orthocentre of a $\Delta ABC$ is $B$ and the circumcentre is $S(a, b)$. If $A$ is the origin,then the coordinates of $C$ are-

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