A triangle has corners at (-1 ,2 ), (3 ,-5 ), and (7 ,4 ). If the triangle is dilated by a factor of 5  about point #(-2 ,6 ), how far will its centroid move?

Apr 26, 2018

Given vertices $\left(a , b\right) , \left(c , d\right) , \left(e , f\right)$, dilation point $\left(p , q\right)$, dilation factor $r$, the distance the centroid moves after dilation is:

$\frac{r - 1}{3} \sqrt{{\left(a + c + e - 3 p\right)}^{2} + {\left(b + d + e - 3 q\right)}^{2}}$

$= \frac{4}{3} \sqrt{{\left(- 1 + 3 + 7 - 3 \left(- 2\right)\right)}^{2} + {\left(2 \pm 5 + 4 - 3 \left(6\right)\right)}^{2}} = \frac{4}{3} \sqrt{514}$

Explanation:

I've answered one or two of these before. Let's do this one in general:

Given a triangle with vertices $\left(a , b\right) , \left(c , d\right) , \left(e , f\right)$ and dilation point $\left(p , q\right)$ determine how far the centroid moves when dilated by a factor of $r$.

The centroid is $\left(u , v\right) = \left(\frac{1}{3} \left(a + c + e\right) , \frac{1}{3} \left(b + d + e\right)\right)$

The distance $s$ from the centroid to the dilation point is before dilation is

$s = \sqrt{{\left(u - p\right)}^{2} + {\left(v - q\right)}^{2}}$

After dilation it will be $r s$ so the total distance moved is:

$m = \left(r - 1\right) s = \left(r - 1\right) \sqrt{{\left(u - p\right)}^{2} + {\left(v - q\right)}^{2}}$

$m = \left(r - 1\right) \sqrt{{\left(\frac{1}{3} \left(a + c + e\right) - p\right)}^{2} + {\left(\frac{1}{3} \left(b + d + e\right) - q\right)}^{2}}$

$m = \frac{r - 1}{3} \sqrt{{\left(a + c + e - 3 p\right)}^{2} + {\left(b + d + e - 3 q\right)}^{2}}$