To find the value of $x$ such that the point $A(x, 1)$ is equidistant from the points $B(-2, -1)$ and $C(1, 2)$, we need to use the distance formula.

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Since $A$ is equidistant from $B$ and $C$, we set the distances from $A$ to $B$ and $A$ to $C$ equal to each other.

$\sqrt{(x - (-2))^2 + (1 - (-1))^2} = \sqrt{(x - 1)^2 + (1 - 2)^2}$

Simplify the expressions inside the square roots:

$\sqrt{(x + 2)^2 + (1 + 1)^2} = \sqrt{(x - 1)^2 + (1 - 2)^2}$

$\sqrt{(x + 2)^2 + 4} = \sqrt{(x - 1)^2 + 1}$

Now, square both sides to eliminate the square roots:

$(x + 2)^2 + 4 = (x - 1)^2 + 1$

Expand both sides:

$(x^2 + 4x + 4) + 4 = (x^2 - 2x + 1) + 1$

Simplify the equation:

$x^2 + 4x + 8 = x^2 - 2x + 2$

Next, subtract $x^2$ from both sides:

$4x + 8 = -2x + 2$

Add $2x$ to both sides:

$6x + 8 = 2$

Subtract 8 from both sides:

$6x = -6$

Finally, divide by 6:

$x = -1$

So, the value of $x$ is $\boxed{-1}$.

Would you like further details, or do you have any questions about this solution?

Here are 5 related questions:

1. How would the solution change if point $A$ were given different coordinates?
2. Can you determine the midpoint of segment $BC$?
3. What is the distance between points $B$ and $C$?
4. How do you derive the distance formula?
5. What is the significance of squaring both sides in distance problems?

Tip: Always verify the solution by plugging the value back into the original equation to ensure it satisfies the condition given in the problem.