# A triangle has corners at (3, 9 ), ( 6, -5), and ( 4, -1). If the triangle is reflected across the x-axis, what will its new centroid be?

Apr 10, 2016

$\left(\frac{13}{3} , - 1\right)$

#### Explanation:

The first step is to find the coordinates of the centroid.

Given the 3 vertices of a triangle $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) , \left({x}_{3} , {y}_{3}\right)$

the x-coord of centroid ${x}_{c} = \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{1}{3} \left({x}_{1} + {x}_{2} + {x}_{3}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

and y-coord of centroid ${y}_{c} = \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{1}{3} \left({y}_{1} + {y}_{2} + {y}_{3}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here let $\left({x}_{1} , {y}_{1}\right) = \left(3 , 9\right) , \left({x}_{2} , {y}_{2}\right) = \left(6 , - 5\right) , \left({x}_{3} , {y}_{3}\right) = \left(4 , - 1\right)$

Hence coords of centroid

$= \left[\frac{1}{3} \left(3 + 6 + 4\right) , \frac{1}{3} \left(9 - 5 - 1\right)\right] = \left(\frac{13}{3} , 1\right)$

Now under reflection in the x-axis a point (x,y) → (x , -y)

new centroid (13/3 , 1) → (13/3 , -1)#