The mid-points of the sides of a triangle are | vedclass

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(A) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.Given mid-points are $(2, 1)$,$(-1, -3)$,and $(4, 5)$.For $x$-co

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$(7, 9), (-3, -7), (1, 1)$

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(A) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.Given mid-points are $(2, 1)$,$(-1, -3)$,and $(4, 5)$.For $x$-coordinates: $\frac{x_1 + x_2}{2} = 2$,$\frac{x_2 + x_3}{2} = -1$,$\frac{x_3 + x_1}{2} = 4$.Adding these equations: $x_1 + x_2 + x_3 = 5$.Subtracting individual equations: $x_3 = 5 - 4 = 1$,$x_1 = 5 - (-2) = 7$,$x_2 = 5 - 8 = -3$.For $y$-coordinates: $\frac{y_1 + y_2}{2} = 1$,$\frac{y_2 + y_3}{2} = -3$,$\frac{y_3 + y_1}{2} = 5$.Adding these equations: $y_1 + y_2 + y_3 = 3$.Subtracting individual equations: $y_3 = 3 - 2 = 1$,$y_1 = 3 - (-6) = 9$,$y_2 = 3 - 10 = -7$.The vertices are $(7, 9), (-3, -7)$,and $(1, 1)$.

The mid-points of the sides of a triangle are $(2, 1)$,$(-1, -3)$,and $(4, 5)$. The coordinates of its vertices are:

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