# A triangle has corners at (4, 6 ), ( 1 , 7), and ( 3, -4). If the triangle is reflected across the x-axis, what will its new centroid be?

May 25, 2016

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

#### Explanation:

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

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

Then the coords of the centroid are found as follows.

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

here let $\left({x}_{1} , {y}_{1}\right) = \left(4 , 6\right) , \left({x}_{2} , {y}_{2}\right) = \left(1 , 7\right) , \left({x}_{3} , {y}_{3}\right) = \left(3 , - 4\right)$

x-coord =1/3(4+1+3)=8/3" #

and y-coord $= \frac{1}{3} \left(6 + 7 - 4\right) = 3$

coords of centroid $= \left(\frac{8}{3} , 3\right)$

now, under a reflection in the x-axis

a point (x ,y) → (x ,-y)

hence coords of centroid $\left(\frac{8}{3} , 3\right) \Rightarrow \left(\frac{8}{3} , - 3\right)$

May 25, 2016

New centroid is $\left(\frac{8}{3} , - 3\right)$

#### Explanation:

The centroid of the triangle with corners at $\left(4 , 6\right)$, $\left(1 , 7\right)$ and $\left(3 , - 4\right)$ is

$\left(\frac{4 + 1 + 3}{3} , \frac{6 + 7 + \left(- 4\right)}{3}\right)$ or $\left(\frac{8}{3} , 3\right)$

When triangle is reflected across the x-axis, its centroid too is reflected across the x-axis

and as reflection of a point $\left({x}_{1} , {y}_{1}\right)$ in x-axis is $\left({x}_{1} , - {y}_{1}\right)$,

New centroid is $\left(\frac{8}{3} , - 3\right)$