# A triangle has corners at #(6 ,9 )#, #(-2 ,-1 )#, and #(1 ,-1 )#. If the triangle is dilated by a factor of #1/3 # about point #(-5 ,-2 ), how far will its centroid move?

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To find the centroid of a triangle, you take the average of the x-coordinates of the vertices to find the x-coordinate of the centroid, and similarly for the y-coordinate.

The centroid of a triangle is given by the coordinates (x_c, y_c), where: [ x_c = \frac{x_1 + x_2 + x_3}{3} ] [ y_c = \frac{y_1 + y_2 + y_3}{3} ]

Given the coordinates of the vertices: [ (6, 9), (-2, -1), (1, -1) ]

We can substitute these values into the formulas to find the centroid of the original triangle.

The x-coordinate of the centroid: [ x_c = \frac{6 + (-2) + 1}{3} = \frac{5}{3} ]

The y-coordinate of the centroid: [ y_c = \frac{9 + (-1) + (-1)}{3} = \frac{7}{3} ]

Now, if we dilate the triangle by a factor of ( \frac{1}{3} ) about the point (-5, -2), we need to find how far the centroid moves.

The distance between the original centroid and (-5, -2) is given by the distance formula: [ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ]

Substituting the coordinates: [ d = \sqrt{(\frac{5}{3} - (-5))^2 + (\frac{7}{3} - (-2))^2} ]

[ d = \sqrt{(\frac{5}{3} + 5)^2 + (\frac{7}{3} + 2)^2} ]

[ d = \sqrt{(\frac{20}{3})^2 + (\frac{13}{3})^2} ]

[ d = \sqrt{\frac{400}{9} + \frac{169}{9}} ]

[ d = \sqrt{\frac{569}{9}} ]

So the distance the centroid moves is ( \frac{\sqrt{569}}{3} ).

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