Given (16,4)(16,4) and (x,−5)(x,−5), find all xx such that the distance between these two points is 1515. Separate multiple answers with a comma.

To find all values of \( x \) such that the distance between the points \( (16, 4) \) and \( (x, -5) \) is 15, we will use the distance formula:

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

Given:
- Point 1: \( (x_1, y_1) = (16, 4) \)
- Point 2: \( (x_2, y_2) = (x, -5) \)
- Distance \( d = 15 \)

Substitute these values into the distance formula:

\[
15 = \sqrt{(x - 16)^2 + (-5 - 4)^2}
\]

Simplify the equation:

\[
15 = \sqrt{(x - 16)^2 + (-9)^2}
\]
\[
15 = \sqrt{(x - 16)^2 + 81}
\]

Square both sides to eliminate the square root:

\[
15^2 = (x - 16)^2 + 81
\]
\[
225 = (x - 16)^2 + 81
\]

Now subtract 81 from both sides:

\[
225 - 81 = (x - 16)^2
\]
\[
144 = (x - 16)^2
\]

Take the square root of both sides:

\[
x - 16 = \pm \sqrt{144}
\]
\[
x - 16 = \pm 12
\]

Solve for \( x \):

1. \( x - 16 = 12 \Rightarrow x = 28 \)
2. \( x - 16 = -12 \Rightarrow x = 4 \)

Thus, the possible values of \( x \) are \( 28 \) and \( 4 \).

### Final answer:
\[
x = 28, 4
\]

Would you like more details or have any questions?

Here are 5 related questions:
1. How does the distance formula work in 3-dimensional space?
2. Can we derive the distance formula from the Pythagorean theorem?
3. How would the problem change if the distance were negative?
4. How do we find the midpoint between two points?
5. What is the slope of the line between the points \( (16, 4) \) and \( (28, -5) \)?

**Tip:** Always double-check your arithmetic when solving equations with squares and square roots.

Given (16,4) and (x,-5), find all x such that the distance between these two points is 15. Separate multiple answers with a comma.

Explore how to find all possible values of x when given the points (16, 4) and (x, -5), with the distance between them being 15. This problem involves using the distance formula, algebraic manipulation, and the Pythagorean theorem. Suitable for students in Grades 9-11.

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