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We are given two points \( A(1, -3) \) and \( B(x, -6) \) on the coordinate plane, with the distance between them being 5 units. We need to determine the possible x-coordinate of point \( B \).

The distance formula between two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) is given by:

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

Substitute the values of \( A(1, -3) \) and \( B(x, -6) \), and set \( d = 5 \):

\[
5 = \sqrt{(x - 1)^2 + (-6 - (-3))^2}
\]

Simplify the expression:

\[
5 = \sqrt{(x - 1)^2 + (-6 + 3)^2}
\]
\[
5 = \sqrt{(x - 1)^2 + (-3)^2}
\]
\[
5 = \sqrt{(x - 1)^2 + 9}
\]

Now square both sides to eliminate the square root:

\[
25 = (x - 1)^2 + 9
\]

Subtract 9 from both sides:

\[
16 = (x - 1)^2
\]

Now take the square root of both sides:

\[
x - 1 = \pm 4
\]

Solve for \( x \):

\[
x = 1 + 4 = 5 \quad \text{or} \quad x = 1 - 4 = -3
\]

Thus, the x-coordinate of point \( B \) can be either \( 5 \) or \( -3 \).

From the answer choices in the image, the correct options are \( 5 \) and \( -3 \). Since only one is needed, you can choose either.

Would you like further clarification or additional details?

Here are 5 related questions:
1. How would the distance formula change in three-dimensional space?
2. Can there be more than two solutions to this problem?
3. What happens if the distance between points \( A \) and \( B \) is less than 5?
4. How do we interpret negative coordinates in a geometric sense?
5. How does changing one coordinate of a point affect the distance between two points?

**Tip:** Always check for both positive and negative solutions when solving quadratic equations!

The distance between points A and B on the coordinate plane is 5 units. If the coordinates of A are (1, -3) and B(x, -6), which of the following can be the x-coordinate of point B?

Learn how to find the x-coordinate of a point on a coordinate plane using the distance formula. This problem involves algebraic concepts, quadratic equations, and the Pythagorean theorem. Suitable for students in Grades 8-10.

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