The point 𝐴(𝑥 ; 1) is equidistant from the points 𝐵(− 2 ; − 1) and 𝐶(1; 2). Determine the value of 𝑥.

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.

Learn how to determine the value of x such that point A(x, 1) is equidistant from points B(-2, -1) and C(1, 2). This problem involves applying the distance formula and solving equations step-by-step. Suitable for Grades 9-12 students.

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