Question

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

Hint:

To find a point equidistant from two given points, use the distance formula and set the distances equal to each other.

Answer 1

Step 1: Let the point on the Y-axis be \( P(0, y) \).}
Let the point \( P(0, y) \) be a point on the Y-axis that is equidistant from points \( A(6, 5) \) and \( B(-4, 3) \).
Step 2: Use the distance formula.}
The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] We are given that the point \( P \) is equidistant from points \( A \) and \( B \), so: \[ \text{Distance from } P \text{ to } A = \text{Distance from } P \text{ to } B \]
Step 3: Set up the distance equation.}
The distance from \( P(0, y) \) to \( A(6, 5) \) is: \[ PA = \sqrt{(6 - 0)^2 + (5 - y)^2} = \sqrt{36 + (5 - y)^2} \] The distance from \( P(0, y) \) to \( B(-4, 3) \) is: \[ PB = \sqrt{(-4 - 0)^2 + (3 - y)^2} = \sqrt{16 + (3 - y)^2} \] Since \( PA = PB \), we can equate the two distances: \[ \sqrt{36 + (5 - y)^2} = \sqrt{16 + (3 - y)^2} \]
Step 4: Solve the equation.}
Square both sides to eliminate the square roots: \[ 36 + (5 - y)^2 = 16 + (3 - y)^2 \] Expand both sides: \[ 36 + (25 - 10y + y^2) = 16 + (9 - 6y + y^2) \] Simplify: \[ 36 + 25 - 10y + y^2 = 16 + 9 - 6y + y^2 \] \[ 61 - 10y = 25 - 6y \] Now, solve for \( y \): \[ 61 - 25 = 10y - 6y \] \[ 36 = 4y \] \[ y = 9 \] % Final Answer Final Answer:
The point on the Y-axis that is equidistant from points \( A(6, 5) \) and \( B(-4, 3) \) is \( P(0, 9) \).