JEE Challenger
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MathEasyAdv 2007 • Paper 2

Centroid of a Triangle Dividing it into Equal Area Subtriangles

Let O(0,0)O(0, 0), P(3,4)P(3, 4), Q(6,0)Q(6, 0) be the vertices of the triangle OPQOPQ. The point RR inside the triangle OPQOPQ is such that the triangles OPROPR, PQRPQR, OQROQR are of equal area. The coordinates of RR are

Options

A

(43,3)\left(\frac{4}{3}, 3\right)

B

(3,23)\left(3, \frac{2}{3}\right)

C

(3,43)\left(3, \frac{4}{3}\right)

Correct
D

(43,23)\left(\frac{4}{3}, \frac{2}{3}\right)

Topics & Concepts

Coordinate GeometryCentroidArea of TriangleTriangles

Step-by-Step Solution

Let R(x,y)R(x, y) be an interior point of △OPQ\triangle OPQ.

A known geometric property of triangles states that the unique interior point which divides a triangle into three sub-triangles of equal area is the centroid (GG) of the triangle:
Area(△OPR)=Area(△PQR)=Area(△OQR)=13Area(△OPQ)  ⟺  R is the centroid G\text{Area}(\triangle OPR) = \text{Area}(\triangle PQR) = \text{Area}(\triangle OQR) = \frac{1}{3}\text{Area}(\triangle OPQ) \iff R \text{ is the centroid } G

The coordinates of the centroid are given by:
x=x1+x2+x33=0+3+63=93=3x = \frac{x_1 + x_2 + x_3}{3} = \frac{0 + 3 + 6}{3} = \frac{9}{3} = 3
y=y1+y2+y33=0+4+03=43y = \frac{y_1 + y_2 + y_3}{3} = \frac{0 + 4 + 0}{3} = \frac{4}{3}

Thus, the coordinates of RR are (3,43)\left(3, \frac{4}{3}\right).

Correct Answer: C

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