Find a point which is equidistant from the points $A(-5, 4)$ and $B(-1, 6)$. How many such points are there?

(D) Let $P(h, k)$ be a point equidistant from $A(-5, 4)$ and $B(-1, 6)$.
By the distance formula,$PA = PB$,which implies $PA^2 = PB^2$.
$(-5 - h)^2 + (4 - k)^2 = (-1 - h)^2 + (6 - k)^2$
Expanding both sides:
$25 + h^2 + 10h + 16 + k^2 - 8k = 1 + h^2 + 2h + 36 + k^2 - 12k$
Simplifying the equation:
$10h - 2h - 8k + 12k + 41 - 37 = 0$
$8h + 4k + 4 = 0$
Dividing by $4$,we get:
$2h + k + 1 = 0$
This is the equation of the perpendicular bisector of the line segment $AB$. Any point lying on this line will be equidistant from $A$ and $B$. Since a line contains an infinite number of points,there are infinitely many such points.
Replacing $h$ and $k$ with $x$ and $y$,the locus of all such points is $2x + y + 1 = 0$.