# A triangle has corners at (6, 4 ), ( 1, -2), and ( 4, -1)#. If the triangle is reflected across the x-axis, what will its new centroid be?

Jul 5, 2016

$\left(\frac{11}{3} , - \frac{1}{3}\right)$

#### Explanation:

The first step is to find the centroid of the given triangle.

If $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) \text{ and } \left({x}_{3} , {y}_{3}\right)$ are the vertices of a triangle,
Then
$\textcolor{red}{\text{-----------------------------------------------------}}$
x-coordinate of centroid $= \frac{1}{3} \left({x}_{1} + {x}_{2} + {x}_{3}\right)$
and y-coordinate of centroid $= \frac{1}{3} \left({y}_{1} + {y}_{2} + {y}_{3}\right)$
$\textcolor{red}{\text{------------------------------------------------------}}$

here the coordinates of the vertices are (6 ,4), (1 ,-2) and (4 ,-1)

x-coordinate $= \frac{1}{3} \left(6 + 1 + 4\right) = \frac{11}{3}$

and y-coordinate $= \frac{1}{3} \left(4 - 2 - 1\right) = \frac{1}{3}$

hence coordinates of centroid $= \left(\frac{11}{3} , \frac{1}{3}\right)$

Under a reflection in the x-axis

a point (x ,y) → (x ,-y)

hence 'new' centroid $= \left(\frac{11}{3} , - \frac{1}{3}\right)$