Question

Let O(0, 0), P(3,4), Q(6, 0) be the vertices of the 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

• A. $(\frac{4}{3},3)$
• B. $(3,\frac{2}{3})$
• C. $(3,\frac{4}{3})$
• D. $(\frac{4}{3},\frac{2}{3})$

Answer 1

The correct option is C $(3,\frac{4}{3})$

Let coordinate of $R=(a,b)$

Given area of triangle OPR, PQR, and OQR are same.

$\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ 3& 4& 1\\ a& b& 1\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}3& 4& 1\\ 6& 0& 1\\ a& b& 1\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ 6& 0& 1\\ a& b& 1\end{array}\right|$
$\Rightarrow 3b-4a=24-4a-3b=6b$.

Solving this equation be get $a=3,b=\frac{4}{3}$