Find a point on the y-axis which is equidistant from the points A(6, 5) and B(– 4, 3).
Step-by-Step Solution
Key Concept: Use the distance formula. For a point P(0, y) on the y‑axis, set the distances PA and PB equal and solve for y.
1. Let the required point be \(P(0, y)\) because it lies on the y‑axis (\(x=0\)).
2. Write the distance from \(P\) to \(A(6,5)\):
$$PA = \sqrt{(0-6)^2 + (y-5)^2} = \sqrt{36 + (y-5)^2}.$$
3. Write the distance from \(P\) to \(B(-4,3)\):
$$PB = \sqrt{(0+4)^2 + (y-3)^2} = \sqrt{16 + (y-3)^2}.$$
4. Since \(P\) is equidistant from \(A\) and \(B\), set \(PA = PB\). Squaring both sides eliminates the square roots:
$$36 + (y-5)^2 = 16 + (y-3)^2.$$
5. Expand the squares:
$$36 + y^2 - 10y + 25 = 16 + y^2 - 6y + 9.$$
6. Cancel \(y^2\) from both sides and simplify:
$$61 - 10y = 25 - 6y.$$
7. Bring the terms containing \(y\) to one side:
$$61 - 25 = -6y + 10y \Rightarrow 36 = 4y.$$
8. Solve for \(y\):
$$y = \frac{36}{4} = 9.$$
9. Therefore the required point on the y‑axis is \(P(0, 9)\).
Correct Answer: (0, 9)