Question

Line passing through $(1,2)$ cuts the axis at $P,Q,$ find slope of the line such that triangle OPQ area is minimum, $O=(0,0)$

Answer 1

Equation of line with $x-$intercept $a$ and $y-$ intercept $b$ is,

$\Rightarrow$ $\frac{x}{a}+\frac{y}{b}=1$

So, $OP=a$ and $OQ=b$

Area of $△OPQ=A=\frac{1}{2}ab$ ----- ( 1 )

Line is passing through $(1,2)$

$\Rightarrow$ $\frac{1}{a}+\frac{2}{b}=1$

$\Rightarrow$ $b=\frac{2a}{a-1}$

$\Rightarrow$ $A=\frac{a^2}{a-1}$ [ From ( 1 ) ]

Differentiating both sides w.r.t $a,$ we get

$\Rightarrow$ $\frac{dA}{da}=-\frac{a^2\frac{d}{da}(a-1)-(a-1)\frac{d}{da}\left(a^2\right)}{(a-1{)}^2}$

$\Rightarrow$ $\frac{dA}{da}=-\frac{a^2\left(1\right)-(a-1)2a}{(a-1{)}^2}$

$\Rightarrow$ $\frac{dA}{da}=\frac{a^2-2a}{(a-1{)}^2}$

To minimize the area, $\frac{da}{da}=0$

$\Rightarrow$ $\frac{dA}{da}=\frac{a^2-2a}{(a-1{)}^2}=0$

$a=0,2$

At $a=0,$ $Area=0$ Not possible

$a=2$ is minima

$\Rightarrow$ $b=\frac{2a}{a-1}=\frac{2\left(2\right)}{2-1}=4$

$\Rightarrow$ $Slope=\frac{-b}{a}=\frac{-4}{2}=-2$