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

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Aicharya · Accepted Answer

$A=(3,1)$ 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.

A straight line \(L_1\) passing through \(A(3,1)\) meets the coordinate axes at \(P\) and \(Q\) such that…

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