Question

find the centroid of the triangle whose vertices are (2,-3),(-4,2) and (8,13).

Answer 1

Given

$\tt \to \: P(2,-3) ,Q(-4,2) \: and \: R(8,13)$

To Find the centroid of the Triangle

Formula

$\tt \to \: G = \bigg(\dfrac{ x_1 +x_2 + x_3 }{3} , \dfrac{y_1 + y_2 + y_3}{3} \bigg)$

Now we have

$\tt \to \: P(x_1 = 2,y_1 = -3) ,Q(x_2 = -4,y_2 = 2) \: and \: R(x_3 = 8,y_3 = 13)$

Now Put the Value on formula

$\tt \to \: G = \bigg( \dfrac{2 - 4 + 8}{3} , \dfrac{ - 3 + 2 + 13}{3} \bigg)$

$\tt \to \: G = \bigg( \dfrac{10 - 4 }{3} , \dfrac{ 10 + 2 }{3} \bigg)$

$\tt \to \: G = \bigg( \dfrac{6 }{3} , \dfrac{ 12}{3} \bigg)$

$\tt \to \: G = \bigg( 2 , 4 \bigg)$

Answer

$\tt \to \: G = \bigg( 2 , 4 \bigg)$

More Formula

1)Distance Formula

$\tt \to \: pq = \sqrt{(x_2 - x_1) {}^{2} + (y_2 - y_1) {}^{2} }$

2) Mid Point

$\tt \to \: p = \bigg( \dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2} \bigg)$