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

Answer 1

Centroid moves by #~~ color(green)(33.44)# due to dilation factor of 5 about point #((5),(-5))#

Centroid #G( ((6+1-2)/3),((5-6+9)/3)) = ((5/3),(8/3))#
Dilated about #((5),(-5))# by a factor of 5

New coordinates of centroid G'

#vec(G'D( = 5 * vec(GD)#

((x-5),(y+5)) -> 5((5/3 - 5),(8/3+5)) = ((50/3),(115/3))#

#G'((x),(y)) = ((15),(100/3))#

Distance between G & G'

#vec(GG') = sqrt((15-5/3)^2 + (100/3-8/3)^2) ~~ color(green)(33.44)#
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Answer 2

To find the centroid of the original triangle, we first calculate the coordinates of the centroid using the formula:

[ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) ]

where ((x_1, y_1)), ((x_2, y_2)), and ((x_3, y_3)) are the coordinates of the triangle's vertices.

Next, we dilate the triangle about point ((5, -5)) by a factor of 5. This means we multiply the coordinates of each vertex of the original triangle by 5, centered at ((5, -5)).

Then, we find the centroid of the dilated triangle using the same formula.

Finally, we find the distance between the centroids of the original and dilated triangles using the distance formula:

[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ]

where ((x_1, y_1)) and ((x_2, y_2)) are the coordinates of the centroids of the original and dilated triangles, respectively.

Let's calculate the centroids and the distance between them:

Centroid of the original triangle: [ x_{\text{centroid}} = \frac{6 + 1 - 2}{3} = \frac{5}{3} ] [ y_{\text{centroid}} = \frac{5 - 6 + 9}{3} = 2 ]

Centroid of the dilated triangle: [ x_{\text{centroid}}' = 5 \times \frac{5}{3} = \frac{25}{3} ] [ y_{\text{centroid}}' = -5 \times 2 = -10 ]

Distance between centroids: [ \text{Distance} = \sqrt{\left(\frac{25}{3} - \frac{5}{3}\right)^2 + (-10 - 2)^2} ] [ = \sqrt{\left(\frac{20}{3}\right)^2 + (-12)^2} ] [ = \sqrt{\frac{400}{9} + 144} ] [ = \sqrt{\frac{400}{9} + \frac{1296}{9}} ] [ = \sqrt{\frac{1696}{9}} ] [ \approx \sqrt{188.44} ] [ \approx 13.74 ]

Therefore, the centroid of the dilated triangle moves approximately 13.74 units from the centroid of the original triangle.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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