What is the locus of point that is equidistant from points P(1,3) and Q(3,5)?

Analyze the question: The locus of points equidistant from two fixed points is the perpendicular bisector of the line joining them. So we need the midpoint of PQ and the line through it that is perpendicular to PQ.

Concept:

• Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2)
• Gradient = (y₂−y₁)/(x₂−x₁)
• Perpendicular gradients multiply to −1

Solution:

1. Midpoint of PQ = ((1+3)/2, (3+5)/2) = (2, 4)

2. Gradient of PQ = (5−3)/(3−1) = 2/2 = 1

3. Gradient of the perpendicular bisector = −1/1 = −1

4. Equation through (2, 4) with gradient −1: y − 4 = −1(x − 2) y = −x + 2 + 4 y = −x + 6

Check using distances: Let (x, y) be equidistant from P and Q: (x−1)² + (y−3)² = (x−3)² + (y−5)² x² − 2x + 1 + y² − 6y + 9 = x² − 6x + 9 + y² − 10y + 25 −2x − 6y + 10 = −6x − 10y + 34 4x + 4y = 24 x + y = 6, so y = −x + 6 ✓

Answer: A. y = −x + 6

Tip: For "equidistant from two points" questions, always think "perpendicular bisector": find the midpoint, flip and negate the gradient, then use y − y₁ = m(x − x₁). You can also quickly test options by checking that the midpoint (2, 4) satisfies the equation: only A gives 4 = −2 + 6.