Question

Find the centroid of the triangle, whose vertices are (-4,4), (-2,2) and (6-6)​

Answer 1

Answer:

$centeriod = x1 + x2 + x3 \div 3 \: \: y1 + y2 + y3 \div 3 \\ = - 4 + ( - 2) + 6 \div 3 \: \: 4 + 2 + ( - 6) \div 3 \\ = 0 \div 3 \: \: \: 0 \div 3 \\ = 0 \: \: 0$

Answer 2

$\red{\bold{\underline{\underline{Answer}}}}$

$\tt\green{\therefore{Centroid=(0,0)}}$

$\pink{\bold{\underline{Step-by-step\:explanation}}}$

• Given

$\tt \implies Coordinate \: of \: p = (-4,4) \\ \\ \tt \implies Coordinate \: of \: q= (-2,2) \\ \\ \tt \implies Coordinate \: of \: r= (6,-6)$

• To find

$\tt \implies Centroid = ?$

For finding value of centroid :

$\tt \implies x = \frac{ x_{1} + x_{2} + x_{3} }{3} \\ \\ \tt \implies x = \frac{-4 -2 +6}{3} \\ \\ \tt \implies x = 0 \\ \\ \tt \implies y=\frac{ y_{1} + y_{2} + y_{3} }{3} \\ \\\tt \implies y= \frac{4 + 2-6}{3} \\ \\ \tt \implies y= 0 \\ \\ \green{\tt \therefore Centroid \: is \: (0,0)}$