$\overline{AB} = \begin{pmatrix}7\\1\end{pmatrix}$. D is the point $(-2,1)$ and E is $(h,6)$.

1. Subtract the coordinates of D from the coordinates of E.

Since $D=(-2,1)$ and $E=(h,6)$,
$\overrightarrow{DE}=\begin{pmatrix}h-(-2)\\6-1\end{pmatrix}=\begin{pmatrix}h+2\\5\end{pmatrix}.$

1. Write the equality of lengths using the magnitudes of the two vectors.

From part (a),
$\overrightarrow{DE}=\begin{pmatrix}h+2\\5\end{pmatrix}$.

Since $DE=AB$,
$\sqrt{(h+2)^2+5^2}=\sqrt{7^2+1^2}.$

2. Solve the resulting equation for $h$.

$\sqrt{(h+2)^2+25}=\sqrt{50}$
$(h+2)^2+25=50$
$(h+2)^2=25$
$h+2=\pm 5$
$h=3 \text{ or } h=-7.$