jmc

The distance between the midpoint of $\overline{AB}$ and an endpoint of $\overline{AB}$ is equal to half of the length of $\overline{AB}$. By the distance formula,

$$\begin{array}{rl}AB& =\sqrt{((t-4)-(-2){)}^2+((-1)-(t+3){)}^2}\\ & =\sqrt{(t-2{)}^2+(t+4{)}^2}\\ & =\sqrt{2t^2+4t+20}\end{array}$$Also, we know that $(AB/2{)}^2=t^2/2⟹AB=2\sqrt{t^2/2}=\sqrt{2t^2}$. Setting these two expressions equal and squaring, we obtain $$A{B}^2=2t^2=2t^2+4t+20⟹4t+20=0.$$Thus, $t=\boxed{-5}$.