Question

A straight line 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 $\triangle OPQ$ is_________ sq. units

• A. $\frac{100}{3}$
• B. $\frac{25}{3}$
• C. $\frac{50}{3}$
• D. $\frac{200}{3}$

Answer 1

Answer: (C)

$A=(3,1)$
Let slope of line be $m$
$\therefore y-y_{1}=m\left(x-x_{1}\right)$ [be the required line]
$y-1=m(x-3)$
$y-1=m x-3 m$
$m x-y+(1-3 m)=0 ...(i)$

The greatest distance of line from origin passes through $A(3,1)$ is
perpendicular to the given line.
$\therefore O A \perp P Q$
Slope of $O A \times$ Slope of $P Q=-1$
$\frac{1}{3} \times m=-1 \Rightarrow m=-3$
Put, $m=-3$ in Eq. (i)
$-3 x-y+(1-3)(-3)=0$
$-3 x-y+(10)=0$
$3 x+y+10=0$
$\therefore $ Area of
$\triangle P O Q=\frac{1}{2}\left|\frac{c^{2}}{a b}\right|$
$=\frac{1}{2}\left|\frac{100}{3 \times 1}\right|=\frac{50}{3}$ sq units