Question

If $O\left(0,0\right),P\left(3,4\right),Q\left(6,0\right)$ be the vertices of $△OPQ$.

The point $R$ inside the $△OPQ$ is such that the triangles $ORP,ORQ,PRQ$ are of equal area.

Then the co-ordinates 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 1

The correct option is C

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

Explanation for the correct option

Step 1: Solve for the co-ordinates of $R$

Given that vertices of $△OPQ$ are $O\left(0,0\right),P\left(3,4\right),Q\left(6,0\right)$ and the areas of triangles $ORP,ORQ,PRQ$ are equal.

Let the co-ordinates of $R$ be $\left(a,b\right)$

We know that area of triangle with co-ordinates $\left(x_1,y_1\right),\left(x_2,y_2\right),\left(x_3,y_3\right)$ is $\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$

Area of $△ORP=△ORQ=△PRQ$ [Given]

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

Step 2: Solve further for the co-ordinates of $R$

Let,

$$\begin{gathered} 4a-3b....\left[1\right] \\ 6b....\left[2\right] \\ 24-4a-3b....\left[3\right] \end{gathered}$$

Solving $\left[1\right]$ and $\left[2\right]$, we get,

$\begin{array}{crcc}\Rightarrow & 4a-3b& =& 6b\\ \Rightarrow & 4a& =& 9b\end{array}$

Solving $\left[2\right]$ and $\left[3\right]$, we get,

$\begin{array}{crcc}\Rightarrow & 24-4a-3b& =& 6b\\ \Rightarrow & 24-9b-3b& =& 6b\\ \Rightarrow & 18b& =& 24\\ \Rightarrow & b& =& \frac{24}{18}\\ \Rightarrow & b& =& \frac{4}{3}\end{array}$

$\begin{array}{crcl}\therefore & 4a& =& 9b\\ \Rightarrow & 4a& =& 9\left(\frac{4}{3}\right)\\ \Rightarrow & a& =& 3\end{array}$

Step 3: Verifying the co-ordinates of $R$

Given that Area of $△ORP=△ORQ=△PRQ$

$\begin{array}{clclcl}\Rightarrow & \left|\begin{array}{ccc}0& 0& 1\\ 3& \frac{4}{3}& 1\\ 3& 4& 1\end{array}\right|& =& \left|\begin{array}{ccc}0& 0& 1\\ 3& \frac{4}{3}& 1\\ 6& 0& 1\end{array}\right|& =& \left|\begin{array}{ccc}3& 4& 1\\ 3& \frac{4}{3}& 1\\ 6& 0& 1\end{array}\right|\\ \Rightarrow & \left|0-0+1\left(12-4\right)\right|& =& \left|0-0+1\left(-8\right)\right|& =& \left|3\left(\frac{4}{3}\right)-4\left(3-6\right)+1\left(-8\right)\right|\\ \Rightarrow & 8& =& 8& =& 8\end{array}$

Therefore, the co-ordinates of $R$ are $\left(3,\frac{4}{3}\right)$

Hence, the correct option is option(C) i.e. $\left(3,\frac{4}{3}\right)$