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

Question

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

Aurora Ayala · Accepted Answer

≈ 6.566 units.

Given # (x_1,y_1),(x_2,y_2)" and " (x_3,y_3)# are the vertices of a triangle then the coordinates of the centroid are. #x_c=1/3(x_1+x_2+x_3)" and " y_c=1/3(y_1+y_2+y_3)# That is the average of the coordinates of the vertices. substituting the given coordinates into the above. #x_c=1/3(1+3+4)=8/3" and " y_c=1/3(9+4+5)=6# #rArr"coordinates of centroid" =(8/3,6)# To calculate the distance from this point to the origin use #color(blue)"Pythagoras' theorem"# #d=sqrt((8/3)^2+6^2)=sqrt(64/9+324/9)# #=sqrt(388/9)≈6.566" to 3 decimal places"#

Mason Adams · Answer

undefined The centroid of a triangle is the point where the three medians intersect. The coordinates of the centroid $G$ of a triangle with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ are given by:

$$G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$$

Using the coordinates provided:
$A(1, 9)$, $B(3, 4)$, and $C(4, 5)$

$$G\left(\frac{1 + 3 + 4}{3}, \frac{9 + 4 + 5}{3}\right)$$

$$G\left(\frac{8}{3}, \frac{18}{3}\right)$$

$$G\left(\frac{8}{3}, 6\right)$$

Now, using the distance formula, the distance between the centroid $G$ and the origin $(0, 0)$ is:

$$d = \sqrt{(x_G - x_0)^2 + (y_G - y_0)^2}$$

$$d = \sqrt{\left(\frac{8}{3} - 0\right)^2 + (6 - 0)^2}$$

$$d = \sqrt{\left(\frac{8}{3}\right)^2 + 6^2}$$

$$d = \sqrt{\frac{64}{9} + 36}$$

$$d = \sqrt{\frac{64}{9} + \frac{324}{9}}$$

$$d = \sqrt{\frac{388}{9}}$$

$$d \approx \frac{\sqrt{388}}{3}$$

Thus, the distance from the centroid of the triangle to the origin is approximately $\frac{\sqrt{388}}{3}$.