# A triangle has corners at (5 ,9 ), (4 ,1 ), and (3 ,8 ). How far is the triangle's centroid from the origin?

Apr 23, 2018

$\textcolor{b l u e}{2 \sqrt{13}}$ units

#### Explanation:

The centroid is the point where the triangles medians meet. A median is a line through a vertex to the midpoint of the opposite side. A triangle has three medians , but will will only need to find two of these to find the point of intersection, which is the centroid.

Chose two sides:

$A B$ and $A C$

Let $A = \left(5 , 9\right) , B = \left(4 , 1\right) , C = \left(3 , 8\right)$

Find the coordinates of the midpoints of these two sides:

The coordinates of the midpoint of a line are given by:

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

For $A B$

$\left(\frac{5 + 4}{2} , \frac{9 + 1}{2}\right) = \left(\frac{9}{2} , 5\right)$

For $A C$

$\left(\frac{5 + 3}{2} , \frac{9 + 8}{2}\right) = \left(4 , \frac{17}{2}\right)$

$A B$ passes through vertex $C = \left(3 , 8\right)$

$A C$ passes through vertex $B = \left(4 , 1\right)$

We now find the equations of two lines using midpoints and vertices.

For $A B$

$\frac{8 - 5}{3 - \frac{9}{2}} = \frac{3}{- \frac{3}{2}} = - 2$

Using point slope form of a line:

$\left({y}_{2} - {y}_{1}\right) = m \left({x}_{2} - {x}_{1}\right)$

$y - 8 = - 2 \left(x - 3\right)$

$y = - 2 x + 14 \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left[1\right]$

For $A C$

$\frac{1 - \frac{17}{2}}{4 - 4} = \frac{- \frac{15}{2}}{0}$( this is undefined and shows we have a vertical line.

$x = 4 \setminus \setminus \setminus \setminus \setminus \left[2\right]$

Solving simultaneously:

$y = - 2 \left(4\right) + 14 \implies y = 6$

Coordinates of centroid:

$\left(4 , 6\right)$

To find the distance from the origin we use the distance formula:

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

$d = \sqrt{{\left(0 - 4\right)}^{2} + {\left(0 - 6\right)}^{2}} = \sqrt{52} = 2 \sqrt{13}$

PLOT: