Triangle ABC has its centroid at point G(4, 5), and vertex A is located at (2, 3). If point D is the midpoint of side BC, what are the coordinates of D?

Correct — A. The centroid G lies on the median AD and splits it AG : GD = 2 : 1, so GD is half of AG.

Step from A to G: (4 − 2, 5 − 3) = (2, 2)
GD = half of that step = (1, 1)
D = G + (1, 1) = (4 + 1, 5 + 1) = (5, 6) → option (a)

Check with the section formula: (A + 2D) ⁄ 3 = ((2 + 10) ⁄ 3, (3 + 12) ⁄ 3) = (4, 5), which is G.

Why the others are wrong

• (b)
  (5, 8) — (5, 8) gets x right but not y. With D = (5, 8), (A + 2D) ⁄ 3 has y = (3 + 16) ⁄ 3 = 19⁄3, not 5. The step from G to D is +1 in y, not +3.
• (c)
  (6, 7) — (6, 7) is 2G − A, the point that makes G the midpoint of AD. The centroid sits two-thirds of the way down the median, so GD is half of AG, not equal to it.
• (d)
  (6, 8) — (6, 8) is A and G added coordinate by coordinate, which is no centroid relation. It fails the check: (A + 2D) ⁄ 3 = (14⁄3, 19⁄3), not (4, 5).

The centroid is where the three medians of a triangle meet. It is the average of the vertices: G = ((x₁ + x₂ + x₃) ⁄ 3, (y₁ + y₂ + y₃) ⁄ 3).

On each median it sits two-thirds of the way from the vertex, so AG : GD = 2 : 1 and G = (A + 2D) ⁄ 3.

Rearranged, D = (3G − A) ⁄ 2. Here that is ((12 − 2) ⁄ 2, (15 − 3) ⁄ 2) = (5, 6), without ever finding B or C.

Key facts

• Centroid of (x₁, y₁), (x₂, y₂), (x₃, y₃) = ((x₁ + x₂ + x₃) ⁄ 3, (y₁ + y₂ + y₃) ⁄ 3).
• The centroid divides each median in the ratio 2 : 1, the longer part next to the vertex.
• Midpoint of the side opposite A: D = (3G − A) ⁄ 2.

Common traps

• Treating G as the midpoint of AD. That gives 2G − A = (6, 7).
• Dropping the weight 2 on D. G = (A + D) ⁄ 3 gives D = (10, 12), which is B + C, not their midpoint.

No directly related past PYQ was found.