all points on the x axis that are 6 units from the point 2, -3 are, write in a ordered pair please

To find all points on the \(x\)-axis that are 6 units away from the point \((2, -3)\), we can use the distance formula:

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

Here, \(d = 6\), \((x_1, y_1) = (2, -3)\), and \((x_2, y_2)\) is any point on the \(x\)-axis. Since all points on the \(x\)-axis have a \(y\)-coordinate of 0, we can set \(y_2 = 0\). Substituting these values into the distance formula:

\[
6 = \sqrt{(x_2 - 2)^2 + (0 - (-3))^2}
\]

Simplifying the expression inside the square root:

\[
6 = \sqrt{(x_2 - 2)^2 + 3^2}
\]

\[
6 = \sqrt{(x_2 - 2)^2 + 9}
\]

Square both sides to eliminate the square root:

\[
36 = (x_2 - 2)^2 + 9
\]

Subtract 9 from both sides:

\[
27 = (x_2 - 2)^2
\]

Now, take the square root of both sides:

\[
x_2 - 2 = \pm \sqrt{27}
\]

\[
x_2 - 2 = \pm 3\sqrt{3}
\]

Solve for \(x_2\):

\[
x_2 = 2 + 3\sqrt{3} \quad \text{or} \quad x_2 = 2 - 3\sqrt{3}
\]

Therefore, the two points on the \(x\)-axis that are 6 units away from the point \((2, -3)\) are:

\[
(2 + 3\sqrt{3}, 0) \quad \text{and} \quad (2 - 3\sqrt{3}, 0)
\]

### Do you want more details or have any questions?

### Related Questions:
1. How do you find the distance between two points in three-dimensional space?
2. How do you find all points that are a fixed distance from a given point in the coordinate plane?
3. What is the equation of a circle in standard form, and how can you derive it?
4. How do you find the midpoint between two points on a coordinate plane?
5. What are the properties of the distance formula in analytic geometry?

### Tip:
When solving distance problems, squaring both sides early can simplify the algebra but remember to consider both the positive and negative roots when solving for unknowns.

All points on the x-axis that are 6 units from the point (2, -3) are, write in an ordered pair please.

Explore how to find points on the x-axis that are 6 units away from the point (2, -3) using the distance formula. This solution demonstrates the application of coordinate geometry and the Pythagorean Theorem, suitable for students in Grades 9-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation