Point P is at an equal distance from the three vertices of a triangle with coordinates A(0, 0), B(4, 0), and C(0, 3). What are the coordinates of point P?

Correct — A. A point equally far from A, B and C lies on the perpendicular bisector of every side.

Bisector of AB, from (0, 0) to (4, 0): the vertical line x = 2
Bisector of AC, from (0, 0) to (0, 3): the horizontal line y = 1.5
Distance check: PA = PB = PC = √(2² + 1.5²) = √6.25 = 2.5

The bisectors meet at P = (2, 1.5) → option (a)

Why the others are wrong

• (b)
  (2, 2) — (2, 2) lies on x = 2 but not on y = 1.5. PC = √(2² + 1²) = √5 ≈ 2.24, while PA = √8 ≈ 2.83, so the distances differ.
• (c)
  (1.5, 2) — (1.5, 2) swaps the two coordinates. PA is still 2.5, but PB = √(2.5² + 2²) = √10.25 ≈ 3.20, so P is not equidistant.
• (d)
  (2.5, 2.5) — (2.5, 2.5) is on neither bisector. PA = √12.5 ≈ 3.54 but PC = √6.5 ≈ 2.55, far from equal.

The point equally far from all three vertices is the circumcentre, where the perpendicular bisectors of the sides meet.

In a right triangle it is the midpoint of the hypotenuse. Here the legs lie along the axes, so ∠A = 90° and BC is the hypotenuse. The midpoint of (4, 0) and (0, 3) is (2, 1.5), and the circumradius is half of BC = 5⁄2 = 2.5.

Key facts

• The circumcentre is equidistant from all three vertices of a triangle.
• In a right triangle the circumcentre is the midpoint of the hypotenuse.
• Midpoint of (x₁, y₁) and (x₂, y₂) = ((x₁ + x₂)⁄2, (y₁ + y₂)⁄2).

Common traps

• Averaging all three vertices, which gives the centroid (4⁄3, 1), not the circumcentre.
• Writing the midpoint of BC as (1.5, 2), with its coordinates swapped.

No directly related past PYQ was found.