https://file.aiquickdraw.com/mathbot/file/1725969330343ayx5sanc.jpg

The problem in the image asks:

**Find a point on the y-axis which is equidistant from the points A(6, 5) and B(-4, 3).**

### Solution:

Let the point on the y-axis be \( P(0, y) \) (since it's on the y-axis, its x-coordinate is 0). The condition is that the distance from \( P \) to \( A(6, 5) \) should be equal to the distance from \( P \) to \( B(-4, 3) \).

Using the distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \):
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

1. Distance between \( P(0, y) \) and \( A(6, 5) \):
\[
d(P, A) = \sqrt{(0 - 6)^2 + (y - 5)^2} = \sqrt{36 + (y - 5)^2}
\]

2. Distance between \( P(0, y) \) and \( B(-4, 3) \):
\[
d(P, B) = \sqrt{(0 - (-4))^2 + (y - 3)^2} = \sqrt{16 + (y - 3)^2}
\]

Since the distances are equal:
\[
\sqrt{36 + (y - 5)^2} = \sqrt{16 + (y - 3)^2}
\]

Squaring both sides:
\[
36 + (y - 5)^2 = 16 + (y - 3)^2
\]

Expanding both sides:
\[
36 + (y^2 - 10y + 25) = 16 + (y^2 - 6y + 9)
\]

Simplifying:
\[
36 + y^2 - 10y + 25 = 16 + y^2 - 6y + 9
\]

\[
y^2 - 10y + 61 = y^2 - 6y + 25
\]

Cancelling out \( y^2 \) from both sides:
\[
-10y + 61 = -6y + 25
\]

Rearranging:
\[
-10y + 6y = 25 - 61
\]

\[
-4y = -36
\]

Dividing by -4:
\[
y = 9
\]

### Final Answer:
The point on the y-axis equidistant from \( A(6, 5) \) and \( B(-4, 3) \) is \( P(0, 9) \).

---

Would you like any additional details or have any further questions?

Here are five questions that expand on this topic:
1. How do you derive the distance formula between two points in a plane?
2. What other methods can be used to find a point equidistant from two given points?
3. Can this problem be extended to finding a point equidistant from three points?
4. How would the solution change if the point were on a different axis or line?
5. What geometric property does the solution point satisfy in relation to the two given points?

**Tip:** When solving geometry problems involving distances, always double-check for simplifications that arise after using the distance formula.

Find a point on the y-axis which is equidistant from the points A(6, 5) and B(-4, 3).

Learn how to find a point on the y-axis equidistant from points A(6, 5) and B(-4, 3) using the distance formula. This solution involves solving an equation derived from equidistance properties in coordinate geometry. 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