# A triangle has corners at (9, 4 ), ( 5, -9), and ( 2, -3)#. If the triangle is reflected across the x-axis, what will its new centroid be?

Jul 10, 2016

$\left(\frac{16}{3} , \frac{8}{3}\right)$

#### Explanation:

The first step is to calculate the coordinates of the existing centroid.
Given the 3 vertices of a triangle

$\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) \text{ and } \left({x}_{3} , {y}_{3}\right)$

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

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

Basically, this is the average of the x and y-coordinates of the vertices.

Thus x-coordinate = $\frac{1}{3} \left(9 + 5 + 2\right) = \frac{16}{3}$

and y-coordinate = $\frac{1}{3} \left(4 - 9 - 3\right) = - \frac{8}{3}$

coordinates of centroid $= \left(\frac{16}{3} , - \frac{8}{3}\right)$

Under a reflection in the x-axis

a point (x ,y) → (x ,-y)

hence $\left(\frac{16}{3} , - \frac{8}{3}\right) \to \left(\frac{16}{3} , \frac{8}{3}\right) \text{ new centroid}$