# A triangle has corners at (1 ,3 ), (9 ,2 ), and (6 ,7 ). How far is the triangle's centroid from the origin?

Mar 2, 2017

$\text{The answer is l=6.67 units}$

#### Explanation:

$\text{the triangle A(1,3),B(9,2),C(6,7) is shown in figure below}$

$\text{The centroid of any triangle can be calculated using the formula below.}$

$c e n t r o i d \left(x , y\right)$

$\text{Centroid of any triangle represents by two values (x,y)}$

$x = \frac{{x}_{A} + {x}_{B} + {x}_{C}}{3}$

$y = \frac{{y}_{A} + {y}_{B} + {y}_{C}}{3}$

$A \left(1 , 3\right) \text{ "rArr " "x_A=1" , } {y}_{A} = 3$

$B \left(9 , 2\right) \text{ "rArr" "x_B=9" , } {y}_{B} = 2$

$C \left(6 , 7\right) \text{ "rArr" "x_C=6" , } {y}_{C} = 7$

$x = \frac{1 + 9 + 6}{3} = \frac{16}{3} = 5.33$

$y = \frac{3 + 2 + 7}{3} = \frac{12}{3} = 4$

$c e n t r o i d \left(5.33 , 4\right)$

$\text{let distance between origin and centroid be l;}$

$l = \sqrt{{\left({y}_{\text{centroid")-y_("origin"))^2+(x_("centroid")-x_("origin}}\right)}^{2}}$

$l = \sqrt{{\left(5.33 - 0\right)}^{2} + {\left(4 - 0\right)}^{2}}$

$l = \sqrt{{\left(5.33\right)}^{2} + {4}^{2}}$

$l = \sqrt{28.41 + 16}$

$l = \sqrt{44.41}$

$l = 6.67$