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.