If (x_i, y_i) are the vertices of an equilate | vedclass

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(C) The given condition $(x_i - 2)^2 + (y_i - 3)^2 = R^2$ implies that all three vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ lie on a circle c

Vedclass · Accepted Answer

$39$

Vedclass · Answer

(C) The given condition $(x_i - 2)^2 + (y_i - 3)^2 = R^2$ implies that all three vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ lie on a circle centered at $(2, 3)$.Since the triangle is equilateral,the circumcenter and the centroid coincide.Thus,the centroid $(G)$ of the triangle is $(2, 3)$.The coordinates of the centroid are given by $G = (\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3})$.Equating the coordinates,we get $\frac{x_1 + x_2 + x_3}{3} = 2 \implies x_1 + x_2 + x_3 = 6$ and $\frac{y_1 + y_2 + y_3}{3} = 3 \implies y_1 + y_2 + y_3 = 9$.Substituting these values into the expression $2(x_1 + x_2 + x_3) + 3(y_1 + y_2 + y_3)$:$= 2(6) + 3(9) = 12 + 27 = 39$.

If $(x_i, y_i)$ are the vertices of an equilateral triangle $ABC$ such that $(x_1 - 2)^2 + (y_1 - 3)^2 = (x_2 - 2)^2 + (y_2 - 3)^2 = (x_3 - 2)^2 + (y_3 - 3)^2$,then find the value of $2(x_1 + x_2 + x_3) + 3(y_1 + y_2 + y_3)$.

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