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. If the area of $\Delta OPQ$ is A sq. units then 3A = ……..

Let m be the slope of the straight line $L_1$

Equation of line $L_1$is $(y-1)=m(x-3)$

$\Rightarrow y-1=mx-3m\Rightarrow mx-y+(1-3m)=0\to \left(1\right)$

Let d be the distance from origin to the line (1)

$\therefore d=\frac{1-3m}{\sqrt{m^2+1}}$

$\Rightarrow d\text{'}=\frac{\sqrt{m^2+1}(-3)-(1-3m)\times \frac{1}{2\sqrt{m^2+1}}\times 2m}{(m^2+1)}=\frac{-3(m^2+1)-m(1-3m)}{{(m^2+1)}^{3/2}}=\frac{-(m+3)}{{(m^2+1)}^{3/2}}$

$d\text{'}=0\Rightarrow m+3=0\Rightarrow m=-3$

Equation of the line L is $(y-1)=-3(x-3)\Rightarrow y-1=-3x+9\Rightarrow 3x+y-10=0$

A = Area of $\Delta OPQ=\frac{c^2}{2\left|ab\right|}=\frac{{(-10)}^2}{2|3\times 1|}=\frac{100}{2\times 3}=\frac{50}{3}\Rightarrow 3A=50$