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What are the coordinates of the centroid of a triangle with vertices P(−4, −1) , Q(2, 2) , and R(2, −3) ?



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Respuesta :

[tex]\bf \qquad \textit{Centroid of a Triangle} \\\\\\ \begin{array}{llll} A(x_1,y_1)\quad B(x_2,y_2)\quad C(x_3,y_3)\\ \quad \\ \left(\cfrac{x_1+x_2+x_3}{3}\quad ,\cfrac{y_1+y_2+y_3}{3}\quad \right) \end{array} \\\\ -------------------------------\\\\ \begin{array}{llll} P(-4,-1)\qquad Q(2,2)\qquad R(2,-3)\\ \quad \\ \left(\cfrac{-4+2+2}{3}\quad ,\cfrac{-1+2-3}{3}\quad \right) \end{array}[/tex]

Answer:

[tex]G(x,y)=(0;-0.67)[/tex]

Step-by-step explanation:

The centroid of a triangle can be calculated with this formula:

[tex]G(x,y)=(\frac{x_{1}+x_{2}+x_{3}}{3}; \frac{y_{1}+y_{2}+y_{3}}{3})[/tex]

The centroid is the average point among the three vertex of a triangle, it's a point formed by the triple interception of all medians of a triangle, the centroid is also called the barycenter or center of mass.

So, replacing all three points into the formula, we have:

[tex]G(x,y)=(\frac{x_{1}+x_{2}+x_{3}}{3}; \frac{y_{1}+y_{2}+y_{3}}{3})\\G(x,y)=(\frac{-4+2+2}{3}; \frac{-1+2+-3}{3})\\G(x,y)=(\frac{0}{3}; \frac{-2}{3})\\G(x,y)=(0;-0.67)[/tex]

Therefore, the centroid is at [tex]G(x,y)=(0;-0.67)[/tex]

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