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Step-by-Step Solution
Step 1: Write the general form of the line
We want the equation of a line passing through the point $(1,\,2)$ with slope $m$.
The slope–intercept form passing through $(x_1,\,y_1)$ is
$$(y - y_1) = m\,(x - x_1).$$
Substituting $x_1 = 1$ and $y_1 = 2$, we get
$$(y - 2) = m\,(x - 1).$$
Step 2: Find the coordinates of $P$ and $Q$ where the line meets the axes
(a) To find $P$ on the $x$-axis, set $y = 0$ in the line equation:
$$
0 - 2 = m\,(x - 1)\quad\Longrightarrow\quad -2 = m\,(x - 1)\quad\Longrightarrow\quad x - 1 = \frac{-2}{m}\quad\Longrightarrow\quad x = 1 - \frac{2}{m}.
$$
Thus, $P\bigl(1 - \tfrac{2}{m},\,0\bigr).$
(b) To find $Q$ on the $y$-axis, set $x = 0$ in the line equation:
$$
y - 2 = m\,(0 - 1)\quad\Longrightarrow\quad y - 2 = -m\quad\Longrightarrow\quad y = 2 - m.
$$
Thus, $Q\bigl(0,\,2 - m\bigr).$
Step 3: Express the area of triangle $OPQ$
The triangle $OPQ$ has vertices $O(0,0)$, $P\bigl(1 - \tfrac{2}{m},\,0\bigr)$, and $Q\bigl(0,\,2 - m\bigr).$
The base can be taken as $OP$ and the height as $OQ$ (since $OP$ lies on the $x$-axis and $OQ$ on the $y$-axis).
So, $OP = \left|1 - \tfrac{2}{m}\right|$ and
$OQ = |2 - m|.$
The area of $\triangle OPQ$ is
$$
\text{Area} = \frac{1}{2}\,\bigl(OP\bigr)\,\bigl(OQ\bigr)
= \frac{1}{2}\,\Bigl(1 - \frac{2}{m}\Bigr)\,\bigl(2 - m\bigr).
$$
(We will work with expressions that treat $m$ in a continuous sense, but keep in mind the absolute values if needed.)
Step 4: Simplify the area expression
Expand:
$$
\frac{1}{2}\Bigl(1 - \frac{2}{m}\Bigr)\,(2 - m)
= \frac{1}{2}\,\biggl[\,2 - m - \frac{4}{m} + \frac{2m}{m}\biggr]
= \frac{1}{2}\,\bigl[4 - m - \frac{4}{m}\bigr]
= 2 - \frac{m}{2} - \frac{2}{m}.
$$
Let
$$
f(m) = 2 - \frac{m}{2} - \frac{2}{m}.
$$
Step 5: Use derivatives to find the minimum area
(a) Compute the first derivative $f'(m)$:
$$
f'(m) = -\frac{1}{2} + \frac{2}{m^2}.
$$
(b) Set $f'(m) = 0$ to locate critical points:
$$
-\frac{1}{2} + \frac{2}{m^2} = 0
\quad\Longrightarrow\quad \frac{2}{m^2} = \frac{1}{2}
\quad\Longrightarrow\quad \frac{1}{m^2} = \frac{1}{4}
\quad\Longrightarrow\quad m^2 = 4
\quad\Longrightarrow\quad m = \pm 2.
$$
(c) Compute the second derivative $f''(m)$ to check which value of $m$ gives a minimum:
$$
f''(m) = \frac{d}{dm}\Bigl(-\frac{1}{2} + \frac{2}{m^2}\Bigr) = -\frac{4}{m^3}.
$$
- For $m = 2$, $f''(2) = -\frac{4}{(2)^3} = -\frac{4}{8} = -\frac{1}{2} < 0,$ which indicates a local maximum.
- For $m = -2$, $f''(-2) = -\frac{4}{(-2)^3} = -\frac{4}{-8} = \frac{1}{2} > 0,$ which indicates a local minimum.
Step 6: Conclude the slope for minimum area
The area is minimized when $m = -2.$ Thus, the slope of the line $PQ$ that results in the least area of $\triangle OPQ$ is $\boxed{-2}.$
