# A triangle has corners at (1, 6 ), ( 1 , 2), and ( 7, -1). If the triangle is reflected across the x-axis, what will its new centroid be?

Jun 17, 2018

$\left(3 , - \frac{7}{3}\right)$.

#### Explanation:

If a point $\left(x , y\right)$ is reflected over the x-axis, then the new point will be $\left(x , - y\right)$. Therefore, instead of first finding the coordinates of the triangle at $\left(1 , 6\right)$,$\left(1 , 2\right)$ and $\left(7 , - 1\right)$ after being reflected across the x-axis, and then finding the centroid of that reflected triangle. We can simply find the centroid of the triangle we are given and then reflect that point across the x-axis.

For information on centroids, visit this webpage.

Now, if we can find out the three medians of the given triangle, and find their point of intersection, we will have the centroid of the triangle. From there, we only need to translate this point from $\left(x , y\right)$ to $\left(x , - y\right)$, and we will arrive at our answer.

The process of finding out the medians of a triangle is quite tedious, so I will only explain how to find out the first. But don't worry, it will become apparent from this one example how to figure out the other two.

We will now figure out the median that originates at point $\left(1 , 6\right)$ and terminates at the midpoint of the line segment that connects points $\left(1 , 2\right)$ and $\left(7 , - 1\right)$. We can determine the midpoint of this line segment using the midpoint formula,

$m i d \left({P}_{1} , {P}_{2}\right) = \left(\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}\right)$
where ${P}_{1}$ and ${P}_{2}$ are two points with coordinates $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ respectively.

Applying this formula to the points ${P}_{1} = \left(1 , 2\right)$ and
${P}_{2} = \left(7 , - 1\right)$, we obtain the midpoint $\left(4 , \frac{1}{2}\right)$.

Now to return to finding the median of the first point of our triangle. We must now write a linear equation for a line that passes through the points $\left(1 , 6\right)$ and the midpoint $\left(4 , \frac{1}{2}\right)$. This can be done thusly:

${f}_{1} \left(x\right) = a x + b$ is the blueprint for our first median. We know:
${f}_{1} \left(1\right) = a \left(1\right) + b = a + b = 6$, and
${f}_{1} \left(4\right) = a \left(4\right) + b = 4 a + b = \frac{1}{2}$.

Solving for the system of linear equations
$a + b = 6$
$4 a + b = \frac{1}{2}$
by means of substitution, we obtain:
$a = - \frac{11}{6}$, and $b = \frac{47}{6}$.

Therefore the linear equation that describes our first median can be written

${f}_{1} \left(x\right) = - \frac{11}{6} x + \frac{47}{6}$.

After finding out the remaining two medians, we set them all equal to find for which $x$ each equation is equal and intersects, and by plugging this $x$ into one of the equations, we will obtain the centroid. Thus:

$- \frac{11}{6} x + \frac{47}{6} = \frac{1}{6} x + \frac{11}{6} = - \frac{5}{6} x + \frac{29}{6}$.

Simplifying and solving, we obtain

$x = 3$.
${f}_{1} \left(3\right) = - \frac{11}{6} \left(3\right) + \frac{47}{6} = \frac{14}{6} = \frac{7}{3.}$

The centroid of the triangle given is therefore $\left(3 , \frac{7}{3}\right)$. Reflected over the x-axis, this point becomes $\left(3 , - \frac{7}{3}\right)$.

Jun 17, 2018

color(blue)((3,-7/3)#

#### Explanation:

If the given triangle is reflected across the x axis, then the centroid will also be reflected in the x axis.

Let the triangle be $A B C$

The co-ordinates of the centroid of a triangle are the arithmetic mean of the x and y co-ordinates of the vertices.

$\therefore$

Centroid is:

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

Centroid of $A B C$

$\left(\frac{1 + 1 + 7}{3} , \frac{6 + 2 - 1}{3}\right) = \left(3 , \frac{7}{3}\right)$

A reflection in the x axis maps:

$\left(x , y\right) \to \left(x , - y\right)$

So image of centroid is:

$\left(x , y\right) \to \left(x , - y\right) = \left(3 , \frac{7}{3}\right) \to \left(3 - \frac{7}{3}\right)$