# A triangle has corners at (3, 5 ), ( 6, 2), and ( 4, 3). If the triangle is reflected across the x-axis, what will its new centroid be?

Apr 22, 2018

$\left(\frac{13}{3} , - \frac{10}{3}\right)$

#### Explanation:

$\text{given the coordinates of the vertices of a triangle}$

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

$\text{then the coordinates of the centroid are}$

•color(white)(x)[1/3(x_1+x_2+x_3),1/3(y_1+y_2+y_3)]

$\text{here } \left({x}_{1} , {y}_{1}\right) = \left(3 , 5\right) , \left({x}_{2} , {y}_{2}\right) = \left(6 , 2\right) , \left({x}_{3} , {y}_{3}\right) = \left(4 , 3\right)$

$\text{hence coordinates of centroid are}$

$\left[\frac{1}{3} \left(3 + 6 + 4\right) , \frac{1}{3} \left(5 + 2 + 3\right)\right] = \left(\frac{13}{3} , \frac{10}{3}\right)$

$\text{under a reflection in the x-axis}$

• " a point "(x,y)to(x,-y)#

$\Rightarrow \left(\frac{13}{3} , \frac{10}{3}\right) \to \left(\frac{13}{3} , - \frac{10}{3}\right) \leftarrow \textcolor{red}{\text{new centroid}}$