Question

Find the centroid of triangle whose vertices are (3,-7) (-8,6)and (12,-3)

Answer 1

The centroid of the triangle whose vertices are
Let,

(x₁,y₁)=(3,-7)
(x₂,y₂)=(-8,6)
(x₃,y₃)=(12,-3)

Then centroid of triangle (g) is :-

$g = ( \frac{x1 + x2 + x3}{3} , \frac{y1 + y2 + y3}{3} )$

$g = ( \frac{3 + ( - 8) + 12}{3} , \frac{ - 7 + 6 + ( - 3)}{3} )$

$g = ( \frac{7}{3} , \frac{ - 4}{3} )$

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{Centroid(G)=}(\frac{7}{3},\frac{-4}{3})}}$

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

$\green{ \underline \bold{Given : }} \\ : \implies \text{Coordinate \: of \: A = (3,-7)} \\ \\ : \implies \text{Coordinate \: of \: B = (-8,6)} \\ \\ : \implies \text{Coordinate \: of \: C = (12,-3)} \\ \\ \red{ \underline \bold{To \: Find : }} \\ : \implies \text{Centroid(G) = ?}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \circ \: \text{Centroid \: of \: triangle(G}) \\ \\ \circ \: \text{For \: x }= \frac{ x_{1} + x_{2} + x_{3} }{3} \\ \\ \circ \: \text{For \: y} = \frac{ y_{1} + y_{2} + y_{3} }{3} \\ \\ \text{Let \: Coordinate \: of \: (g) =( x,y) } \\ \\ \bold{For \: x}\\ : \implies x = \frac{ x_{1} + x_{2} + x_{3} }{3} \\ \\ : \implies x = \frac{3+(-8) + 12}{3} \\ \\ : \implies x = \frac{15-8}{3} \\ \\ \green{: \implies x =\frac{7}{3}} \\ \\ \bold{For \: y}\\ : \implies y= \frac{ y_{1} + y_{2} + y_{3} }{3} \\ \\ : \implies y= \frac{-7+6+(-3)}{3} \\ \\ : \implies y = \frac{6-10}{3} \\ \\ \green{: \implies y =\frac{-4}{3}} \\ \\ \green{\therefore \text{Coordinate \: of \: centroid(G) = }(\frac{7}{3},\frac{-4}{3})}$