find a point on the x-axis which is equidistant from (2,-5) (-2,9)

To find the point on the \(x\)-axis that is equidistant from the points \((2, -5)\) and \((-2, 9)\), let the required point be \((x, 0)\) (since any point on the \(x\)-axis has a \(y\)-coordinate of 0).

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

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

We need to find \(x\) such that the distance from \((x, 0)\) to \((2, -5)\) is equal to the distance from \((x, 0)\) to \((-2, 9)\). Set up the two distances:

1. Distance from \((x, 0)\) to \((2, -5)\):

\[
d_1 = \sqrt{(x - 2)^2 + (0 - (-5))^2} = \sqrt{(x - 2)^2 + 25}
\]

2. Distance from \((x, 0)\) to \((-2, 9)\):

\[
d_2 = \sqrt{(x + 2)^2 + (0 - 9)^2} = \sqrt{(x + 2)^2 + 81}
\]

Since the distances are equal, set them equal to each other:

\[
\sqrt{(x - 2)^2 + 25} = \sqrt{(x + 2)^2 + 81}
\]

Now, square both sides to eliminate the square roots:

\[
(x - 2)^2 + 25 = (x + 2)^2 + 81
\]

Expand both sides:

\[
(x^2 - 4x + 4) + 25 = (x^2 + 4x + 4) + 81
\]

Simplify:

\[
x^2 - 4x + 29 = x^2 + 4x + 85
\]

Now subtract \(x^2\) from both sides:

\[
-4x + 29 = 4x + 85
\]

Combine like terms:

\[
-4x - 4x = 85 - 29
\]

\[
-8x = 56
\]

Solve for \(x\):

\[
x = \frac{56}{-8} = -7
\]

Thus, the point on the \(x\)-axis that is equidistant from \((2, -5)\) and \((-2, 9)\) is \((-7, 0)\).

Would you like more details on this, or do you have any other questions?

Here are some related questions to consider:
1. What is the general formula to find points equidistant from two points?
2. How can you find the perpendicular bisector of the line segment between two points?
3. How do we apply the distance formula in 3-dimensional geometry?
4. Can this problem be solved using the midpoint formula?
5. How does symmetry play a role in problems involving equidistance?

**Tip:** Always remember to check your solution by plugging it back into the original equation to confirm its correctness!

Learn how to find the point on the x-axis that is equidistant from the points (2, -5) and (-2, 9) using the distance formula. This problem involves coordinate geometry, symmetry, and solving quadratic equations. Suitable for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation