Question

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

Hint:

The locus of points equidistant from two fixed points (A and B) is the perpendicular bisector of the line segment AB. This question is asking for the intersection of the y-axis and the perpendicular bisector of the segment joining (6,5) and (-4,3).

Answer 1

Step 1: Understanding the Concept:
Any point on the y-axis has its x-coordinate equal to zero. The problem requires us to find a point P(0, y) that is equidistant from two given points A(6, 5) and B(-4, 3). This means the distance PA must be equal to the distance PB.

Step 2: Key Formula or Approach:
We will use the distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\), which is \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\). To simplify calculations, we can work with the squares of the distances: \(PA^2 = PB^2\).

Step 3: Detailed Explanation:
Let the point on the y-axis be P(0, y).
Let the given points be A(6, 5) and B(-4, 3).
The condition is PA = PB, which implies \(PA^2 = PB^2\).
Using the distance formula for \(PA^2\): \[ PA^2 = (6 - 0)^2 + (5 - y)^2 = 6^2 + (5 - y)^2 = 36 + 25 - 10y + y^2 = 61 - 10y + y^2 \] Using the distance formula for \(PB^2\): \[ PB^2 = (-4 - 0)^2 + (3 - y)^2 = (-4)^2 + (3 - y)^2 = 16 + 9 - 6y + y^2 = 25 - 6y + y^2 \] Now, set \(PA^2 = PB^2\): \[ 61 - 10y + y^2 = 25 - 6y + y^2 \] The \(y^2\) terms on both sides cancel out. \[ 61 - 10y = 25 - 6y \] Rearrange the equation to solve for y: \[ 61 - 25 = -6y + 10y \] \[ 36 = 4y \] \[ y = \frac{36}{4} = 9 \] The point on the y-axis is (0, 9).

Step 4: Final Answer:
The required point on the y-axis is (0, 9).