The medians of a triangle meet at $(0, -3)$ and its two vertices are at $(-1, 4)$ and $(5, 2)$. Then the third vertex is

Given vertices $(-1, 4)$ and $(5, 2)$ the median meeta at centroid $(0,-3)$

we know that in a triangle with vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ then the centroid is $\left(\dfrac {x_1+x_2+x_3}{3},\dfrac {y_1+y_2+y_3}{3}\right)$

Let the third vertex be $(a,b)$

we have,

$(0,-3)=\left ( \dfrac{-1+5+a}{3},\dfrac{4+2+b}{3} \right )$

$0=\dfrac{-1+5+a}{3}\Rightarrow 0=\dfrac{4+a}{3}\Rightarrow 4+a=0\Rightarrow a=-4$

$-3=\dfrac{4+2+b}{3}\Rightarrow -3=\dfrac{6+b}{3}\Rightarrow -9=6+b\Rightarrow b=-15$

$\therefore (-4,-15)$ are the third vertex