If $(2,3)$ is the centroid of the triangle and two vertices are $(4,5)$ and $(-3,5)$, then third vertex is

$\begin{array}{l}\text{Centroid\hspace{0.17em}}G=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})\\ (2,3)=(\frac{4-3+x_3}{3},\frac{5+5+y_3}{3})\\ (2,3)=(\frac{1+x_3}{3},\frac{10+y_3}{3})\\ 2=\frac{1+x_3}{3},3=\frac{10+y_3}{3}\\ 1+x_3=6,10+y_3=9\\ x_3=5,y_3=-1\\ \therefore \text{\hspace{0.17em}Third vertex\hspace{0.17em}}(x_3,y_3)=(5,-1)\end{array}$