# A triangle has corners at (3, 4 ), ( 4, -2), and ( 7, -1). If the triangle is reflected across the x-axis, what will its new centroid be?

Apr 21, 2018

$\left(\frac{14}{3} , - \frac{1}{3}\right)$

#### Explanation:

The centroid is the center of mass, given by the average of the coordinates:

$C = \setminus \frac{1}{3} \left(\left(3 , 4\right) + \left(4 , - 2\right) + \left(7 , - 1\right)\right) = \left(\frac{14}{3} , \frac{1}{3}\right)$

Reflecting through the x axis keeps the x coordinate the same and negates the y coordinate, so we get

$C ' = \left(\frac{14}{3} , - \frac{1}{3}\right)$

Apr 21, 2018

$\left(\frac{14}{3} , - \frac{1}{3}\right)$

#### Explanation:

$\text{given the vertices of a triangle, say}$

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

$\text{coordinates of centroid } = \left[\frac{1}{3} \left({x}_{1} + {x}_{2} + {x}_{3}\right) , \frac{1}{3} \left({y}_{1} + {y}_{2} + {y}_{3}\right)\right]$

$\Rightarrow \text{coordinates } = \left[\frac{1}{3} \left(3 + 4 + 7\right) , \frac{1}{3} \left(4 - 2 - 1\right)\right]$

color(white)(rArr"coordinates ")=(14/3,1/3)larrcolor(blue)"centroid"

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

• " a point "(x,y)to(x,-y)#

$\Rightarrow \left(\frac{14}{3} , \frac{1}{3}\right) \to \left(\frac{14}{3} , - \frac{1}{3}\right) \leftarrow \textcolor{red}{\text{new centroid}}$