Question

Find the Centroid of a ΔPQR,whose vertices are P(1,1) Q(2,2) R(-3,-3) ​

Answer 1

SOLUTION

The given points are P(1,1),Q(2,2) and R(-3,3)

To finD

Centroid of the traingle

Centroid of a triangle is given as :

$\sf S(x,y) = \bigg( \dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3} \bigg)$

Putting the values,we get :

$\implies \sf s(x,y) = \bigg( \dfrac{1 + 2 - 3}{3} , \dfrac{1 + 2 - 3}{3} \bigg) \\ \\ \implies \: \boxed{ \boxed{ \sf{s(x,y) = (0,0)}}}$

The co-ordinates of the centroid are (0,0)

Answer 2

Question:

Find the Centroid of a ΔPQR,whose vertices are P(1,1) Q(2,2) R(-3,-3) .

Answer:

Let (x,y) be the centroid of the triangle whose vertices are (1,1),(2,2),(-3,-3).

$x = \frac{ x_{1} +x_{2} +x_{3}}{3}...........(1) \\ \\ where \\ \implies\: x_{1} = 1 \\ \implies \: x_{2} = 2 \\ \implies \: x_{3} = - 3 \\ \\ put \: value \: x_{1} \: x_{2} \: x_{3} \: in \: given \: equation(1) \\ \\ x = \frac{1 + 2 + ( - 3)}{3} \\ \\ x = \frac{3 - 3}{3} \\ \\ x = \frac{0}{3} \\ \\ x = 0$

Now,

$y = \frac{y_{1} +y_{2} +y_{3}}{3}.........(2) \\ \\ where \\ \implies \: y_{1} = 1\\ \implies \: y_{2} = 2 \\ \implies \: y_{3} = - 3 \\ \\ put \: the \: value \: of \: y_{1} \: y_{2} \: y_{3} \: in \: given \: equation(2) \\ y = \frac{1 + 2 - 3}{3} \\ \\ y = \frac{0}{3} \\ \\ y = 0$

centriod (0,0)