Question

A line is drawn through the point $(1,2)$ to meet the coordinate axes at $P$ and $Q$ such that it forms a triangle $OPQ$, where $O$ is the origin, if the area of the $OPQ$ is least, then the slope of the line $PQ$ is?

Answer 1

Step 1: Find the equation of line $PQ$ in intercept form.

Draw a figure based on given data as shown.

Let $m$ be the slope of the line $PQ$

Then, the equation of $PQ$ using the given point $(1,2)$ and slope $m$ is

$y-2=m(x-1)$

$$\begin{gathered} \Rightarrow y-2=mx-m \\ \Rightarrow mx-y=m-2 \\ \Rightarrow \frac{x}{\left({\displaystyle \frac{m-2}{m}}\right)}-\frac{y}{\left(m-2\right)}=1 \end{gathered}$$

Step 2: Find the Area of $\Delta OPQ$

From the figure,

$PQ$ meets $x$-axis at $P\left(\frac{m-2}{m},0\right)$ and $y$-axis at $Q\left(0,2-m\right)$

$x$-intercept$=$$OP=\frac{m-2}{m}$

$y$-intercept$=$$OQ=2-m$

Area of $\Delta OPQ=\frac{1}{2}\left|OP\times OQ\right|$

$$\begin{gathered} =\frac{1}{2}\left|\frac{m-2}{m}\times 2-m\right| \\ =\frac{1}{2}\left|\left(1-\frac{2}{m}\right)\times \left(2-m\right)\right| \\ =\frac{1}{2}\left|2-m-\frac{4}{m}+2\right| \\ =\frac{1}{2}\left|4-\left(m+\frac{4}{m}\right)\right| \end{gathered}$$

Step 3: Find the value of $m$

Let $f\left(m\right)=4-\left(m+\frac{4}{m}\right)$

Differentiate with respect to $m$.

$\Rightarrow f\text{'}\left(m\right)=-1-\frac{4}{m^2}$

The area of the $OPQ$ is least.

Therefore $f\text{'}\left(m\right)=0$

$$\begin{gathered} \Rightarrow -1+\frac{4}{m^2}=0 \\ \Rightarrow m^2=4 \\ \Rightarrow m=2orm=-2 \end{gathered}$$

If $m=2$the area is $0$ which is not possible.

Therefore, the required slope of $PQ$ is $-2.$