Question

Two vertices of a triangle are (3,2)and (-2,1) and its centroid is (5/3,-1/3) then find the third vertex of the triangle​

Answer 1

Given :-

• Two vertices of triangle are ( 3 , 2 ) and ( -2 , 1 ).

• Coordinates of centroid = ( 5/3 , -1/3 ).

To Find :-

• Coordinates of third vertex.

Solution :-

Let coordinates of third vertex be $\sf (x_3 ,y_3)$ .

$\underline{\boxed{\bf Centroid \ formula = \left( \dfrac{x_1+x_2+x_3}{3} \ , \ \dfrac{y_1+y_2+y_3}{3} \right)} }$

Here :-

$\bullet \sf \ x_1= 3 \\\\\bullet \sf \ x_2 = -2 \\\\\bullet \sf \ y_1 = 2 \\\\\bullet \sf \ y_2 = 1$

$\longrightarrow \sf \dfrac{x_1+x_2+x_3}{3} = \dfrac{5}{3} \\\\\longrightarrow \dfrac{3-2+x_3}{3} = \dfrac{5}{3} \\\\\longrightarrow \dfrac{1+x_3}{3} = \dfrac{5}{3} \\\\\longrightarrow 3(1+x_3) = 15 \\\\\longrightarrow 3+3x_3 = 15 \\\\\longrightarrow 3x_3 = 12 \\\\\longrightarrow \bf x_3 = 4$

$\longrightarrow \sf \dfrac{y_1+y_2+y_3}{3} = \dfrac{-1}{3} \\\\\longrightarrow \dfrac{2+1+y_3}{3} = \dfrac{-1}{3} \\\\\longrightarrow \dfrac{3+y_3}{3} = \dfrac{-1}{3} \\\\\longrightarrow 3(3+y_3) = -3 \\\\\longrightarrow 9+3y_3 = -3 \\\\\longrightarrow 3y_3 = -12 \\\\\longrightarrow \bf y_3 = -4$

Coordinates of the third vertex = ( 4 , -4 )