First slide

Cartesian plane

Question

Let O(0, 0), P(3, 4), and Q(6,0) be the vertices of triangle OPQ. The point R inside the triangle OPQ is such that the triangles OPR, PQR, OQR are of equal area. The coordinates of R are

Easy

A

(4/3,3)
B

(3,2/3)
C

(3,4/3)
D

(4/3,2/3)

Solution

$\begin{aligned} Ar (Δ O P R) = Ar (Δ P Q R) \\ = Ar (Δ O Q R) \end{aligned}$
$R should be the centroid of △ P Q O$ whose coordinates are
$(\frac{3 + 6 + 0}{3}, \frac{4 + 0 + 0}{3}) \equiv (3, \frac{4}{3})$

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