Question
Let $$O\left( {0,\,0} \right),\,P\left( {3,\,4} \right),\,Q\left( {6,\,0} \right)$$ be the vertices of the triangles $$OPQ.$$ The point $$R$$ inside the triangle $$OPQ$$ is such that the triangles $$OPR, \,PQR, \,OQR$$ are of equal area. The coordinates of $$R$$ are-
A.
$$\left( {\frac{4}{3},\,3} \right)$$
B.
$$\left( {3,\,\frac{2}{3}} \right)$$
C.
$$\left( {3,\,\frac{4}{3}} \right)$$
D.
$$\left( {\frac{4}{3},\,\frac{2}{3}} \right)$$
Answer :
$$\left( {3,\,\frac{4}{3}} \right)$$
Solution :
$$\because Ar\left( {\Delta OPR} \right) = Ar\left( {\Delta PQR} \right) = Ar\left( {\Delta OQR} \right)$$
$$\therefore $$ By simply geometry $$R$$ should be the centroid of $$\Delta PQO$$
$$ \Rightarrow R\left( {\frac{{3 + 6 + 0}}{3},\,\frac{{4 + 0 + 0}}{3}} \right) = \left( {3,\,\frac{4}{3}} \right)$$
$$\because Ar\left( {\Delta OPR} \right) = Ar\left( {\Delta PQR} \right) = Ar\left( {\Delta OQR} \right)$$
$$\therefore $$ By simply geometry $$R$$ should be the centroid of $$\Delta PQO$$
$$ \Rightarrow R\left( {\frac{{3 + 6 + 0}}{3},\,\frac{{4 + 0 + 0}}{3}} \right) = \left( {3,\,\frac{4}{3}} \right)$$