Question

If the point P (x, y) is equidistant from the points (3, 6) and (-3, 4), obtain the relation between x and y. Hence, find the coordinates of point P if it lies on x-axis.

Hint:

Whenever a question says a point lies on the x-axis, immediately set \(y = 0\). This reduces the number of variables you need to solve for.

Answer 1

Step 1: Understanding the Concept:
"Equidistant" means the distance from \(P\) to point \(A\) is equal to the distance from \(P\) to point \(B\). We use the distance formula.
Step 2: Key Formula or Approach:
\(PA^2 = PB^2\) \((x - x_1)^2 + (y - y_1)^2 = (x - x_2)^2 + (y - y_2)^2\)
Step 3: Detailed Explanation:
1. Let \(A = (3, 6)\) and \(B = (-3, 4)\).
\[ (x - 3)^2 + (y - 6)^2 = (x - (-3))^2 + (y - 4)^2 \]
2. Expand both sides:
\[ x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16 \] 3. Cancel \(x^2, y^2, \text{ and } 9\) from both sides:
\[ -6x - 12y + 36 = 6x - 8y + 16 \] 4. Rearrange terms:
\[ 12x + 4y = 20 \implies 3x + y = 5 \] 5. If P lies on x-axis: Then \(y = 0\).
6. Substitute \(y=0\) into the relation: \(3x + 0 = 5 \implies x = 5/3\).
Step 4: Final Answer:
Relation: \(3x + y = 5\). Point P on x-axis: \((5/3, 0)\).