Let the vertex A of a triangle ABC be (1, 2), and the mid-point of the side AB be $(5, - 1)$. If the

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Let the vertex A of a triangle ABC be (1, 2), and the mid-point of the side AB be $(5, - 1)$. If the centroid of this triangle is $(3,4)$ and its circumcenter is $(\alpha,\beta)$, then $21(\alpha + \beta)$ is equal to:

Accepted Answer

Sol. [Key Concept: Recover B, C from the midpoint and centroid; get the circumcentre from perpendicular bisectors.]equation of PM : $4x - 3y = 23$equation of PM : $5x - 9y = - 25$$${P\left( \frac{94}{7},\frac{215}{21} \right) }{21(\alpha + \beta) = 497 }$$

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