Question

find the centroid of the triangle whose vertices are (2,-5),(5,11), and (9,9)​

Answer 1

Step-by-step explanation:

$\bf \huge \bigstar \underline{ \blue{AnswEr:-}}$

$\huge\bf† \underline \red{Given :}†$

$\sf \green {‡ \: \:( x_1 \: \: y_1)= (2 \: \: - 5)}$

$\sf \green {‡ \: \:( x_2 \: \: y_2)= (5 \: \: 11)}$

$\sf \green {‡ \: \:( x_3 \: \: y_3)= (9 \: \: 9)}$

$\bf \huge \: † \: \underline \red{Formula \: \: Used} \: †$

$\huge \bf »C = \left (\dfrac{x_1 + x_2 + x_3}{3} \right)\left (\dfrac{y_1 + y_2 + y_3}{3} \right)$

where ,

$\sf \large \: →C \: \: is \: \: a \: \: centroid \: \: of \: \: the \: \: \triangle$

$→ \sf \large \: {x_1 + x_2 + x_3} \: \: are \: \: the \: \: x \: \: coordinates \: \: of \: \: the \: \: vertices \: \: of \: \triangle$

$→ \sf \large \: {y_1 + y_2 + y_3} \: \: are \: \: the \: \: x \: \: coordinates \: \: of \: \: the \: \: vertices \: \: of \: \triangle$

$\huge \leadsto \tt \: C = \frac{2 + 5 + 9}{3} , \frac{ - 5 + 11 + 9}{3}$

$\huge \leadsto \tt \: C = \frac{16}{3} ,5$

$\huge \therefore \tt \purple{Centroid \: \: of \: \: the \: \triangle = \: \frac{16}{3} ,5}$