Two vertices of a triangle are (3 -5) and (-7 4)

Let the coordinates of third vertex be (x, y) and other two vertices are (3, -5) and (-7, 4) and centroid = (2, -1).

Coordinates of Centroid of the triangle are given by

(x1+x2+x33,y1+y2+y33)⇒2=3+(−7)+x3 and −1=−5+4+y3⇒6=x−4 and −3=y−1⇒x=6+4 and y=−3+1⇒x=10 and y=−2.\Big(\dfrac{x1 + x2 + x3}{3}, \dfrac{y1 + y2 + y3}{3}\Big) \\[1em] \Rightarrow 2 = \dfrac{3 + (-7) + x}{3} \text{ and } -1 = \dfrac{-5 + 4 + y}{3} \\[1em] \Rightarrow 6 = x - 4 \text{ and } -3 = y - 1 \\[1em] \Rightarrow x = 6 + 4 \text{ and } y = -3 + 1 \\[1em] \Rightarrow x = 10 \text{ and } y = -2.(3x1​+x2​+x3​​,3y1​+y2​+y3​​)⇒2=33+(−7)+x​ and −1=3−5+4+y​⇒6=x−4 and −3=y−1⇒x=6+4 and y=−3+1⇒x=10 and y=−2.

Hence, the coordinates of centroid are (10, -2).