Question

The points (2, 3) (x, y) and (3, -2) are verticies of a triangle. If origin is

the centroid of this triangle then find (x, y)​

Answer 1

$\bf \underline{ \underline{\maltese\:Given} }$

$\sf The \: points \: (2, 3) \: (x, y) \: and \: (3, -2) \: are \: verticies \: of \: a \: triangle.$

$\sf Origin \: is \: centroid \: of \: triangle.$

$\bf \underline{ \underline{\maltese\:To \: find } }$

$\sf \implies Value \: of \: (x, y) = \: ?$

$\bf \underline{ \underline{\maltese\:Solution} }$

$\sf \underline{As \: we \: know \: that},$

$\sf \implies Points \: of \: origin \: is \: always \: (0,0)$

$\bf \underline{Now},$

$\underline{ \boxed{ \sf Centroid \: formula = (0,0) = \dfrac{(x_1 + x_2 + x_3)}{3} \: , \: \dfrac{(y_1 + y_2 + y_3)}{3}}}$

$\bf \underline{Where},$

$\sf \implies x_1 = 2$

$\sf \implies x_2 = x$

$\sf \implies x_3 = 3$

$\sf \implies y_1 = 3$

$\sf \implies y_2 = y$

$\sf \implies y_3 = - 2$

$\sf \underline{Substituting \: the \: values},$

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

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

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

$\sf \implies(0,0) = \dfrac{5 + x}{3} \: , \: \dfrac{1 + y}{3}$

$\bf \underline{Now},$

$\sf \implies\dfrac{5 + x}{3} = 0 \: , \: \dfrac{1 + y}{3} = 0$

$\sf \implies5 + x = 0 \: , \: 1 + y = 0$

$\sf \implies x = - 5 \: , \: y = - 1$

$\bf \underline{Therefore},$

$\sf \implies Value \: of \: x \: is \: -5$

$\sf \implies Value \: of \: y \: is \: -1$

$\underline{ \boxed{ \bf \red{Hence, \: (x,y) \: = \: (-5,-1)}}}$