A straight line \(L_1\) passing through \(A(3,1)\) meets the coordinate axes at \(P\) and \(Q\) such that its distance from the origin \(O\) is maximum. Then area of \(\triangle O P Q\) is sq. units

• A

  \(\frac{100}{3}\)
• B

  \(\frac{25}{3}\)
• Correct answer
  C

  \(\frac{50}{3}\)
• D

  \(\frac{200}{3}\)

Solution

\(A=(3,1)\) Let slope of line be \(m\) \(\begin{array}{ll} \therefore \quad & y-y_1=m\left(x-x_1\right) \quad \text { [be the required line] } \\ & y-1=m(x-3) \end{array}\) \(\begin{aligned} & y-1=m x-3 m \\ & m x-y+(1-3 m)=0 \quad \ldots (i) \end{aligned}\)
The greatest distance of line from origin passes through \(A(3,1)\) is perpendicular to the given line. \(\therefore \quad O A \perp P Q\) Slope of \(O A \times\) Slope of \(\mathrm{PQ}=-1\) \(\frac{1}{3} \times m=-1 \Rightarrow m=-3\) Put, \(m=-3\) in Eq. (i), \(\begin{aligned} -3 x-y+(1-3)(-3) & =0 \\ -3 x-y+(10) & =0 \\ 3 x+y+10 & =0 \end{aligned}\) \(\therefore\) Area of \(\triangle P O Q=\frac{1}{2}\left|\frac{c^2}{a b}\right|=\frac{1}{2}\left|\frac{100}{3 \times 1}\right|=\frac{50}{3} \mathrm{sq}\) units. Hence, option (c) is correct.

Hence, the correct answer is option: 'C'.