A triangle has corners at $(4, 1)$ $(4, 1 )$ , ( 6, -8) $, and$ $, and$ (3, -2 )#. If the triangle is reflected across the x-axis, what will its new centroid be?

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Answer 1 · 2016-05-25T09:08:21

New centroid is $(\frac{13}{3}, 3)$ $(13/3,3)$

The centroid of the triangle with corners at $(4, 1)$ $(4,1)$ , $(6, - 8)$ $(6,-8)$ and $(3, - 2)$ $(3, -2 )$ is

$(\frac{4 + 6 + 3}{3}, \frac{1 + (- 8) + (- 2)}{3})$ $((4+6+3)/3,(1+(-8)+(-2))/3)$ or $(\frac{13}{3}, - 3)$ $(13/3,-3)$

When triangle is reflected across the x-axis, its centroid too is reflected across the x-axis

and as reflection of a point $(x_{1}, y_{1})$ $(x_1,y_1)$ in x-axis is $(x_{1}, - y_{1})$ $(x_1,-y_1)$ ,

New centroid is $(\frac{13}{3}, 3)$ $(13/3,3)$

A triangle has corners at (4,1)(4, 1 ), ( 6, -8),and, and (3, -2 )#. If the triangle is reflected across the x-axis, what will its new centroid be?