Key Concepts and Formulas

• Equation of a Line (Point-Slope Form): $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and $m$ is the slope.
• Area of a Triangle: Area = $\frac{1}{2} \cdot \text{base} \cdot \text{height}$. For a triangle formed by the origin and the intercepts on the axes, the area is $\frac{1}{2} |x\text{-intercept} \cdot y\text{-intercept}|$.
• Application of Derivatives: To find the minimum or maximum of a function, we find the critical points by setting the first derivative equal to zero and then check the second derivative to confirm whether it's a minimum or maximum.

Step-by-Step Solution

Step 1: Define the line's equation. Since the line passes through the point $(1, 2)$ and has a slope $m$, we can write its equation using the point-slope form: $y - 2 = m(x - 1)$ This is the equation of the line PQ.

Step 2: Find the x and y intercepts. To find the x-intercept (point P), we set $y = 0$: $0 - 2 = m(x - 1)$ $-2 = mx - m$ $mx = m - 2$ $x = \frac{m - 2}{m} = 1 - \frac{2}{m}$ So, the x-intercept is $P = \left(1 - \frac{2}{m}, 0\right)$.

To find the y-intercept (point Q), we set $x = 0$: $y - 2 = m(0 - 1)$ $y - 2 = -m$ $y = 2 - m$ So, the y-intercept is $Q = (0, 2 - m)$.

Step 3: Express the area of triangle OPQ in terms of m. The area of triangle $OPQ$ is given by: $A = \frac{1}{2} |x\text{-intercept} \cdot y\text{-intercept}|$ $A = \frac{1}{2} \left| \left(1 - \frac{2}{m}\right)(2 - m) \right|$ $A = \frac{1}{2} \left| 2 - m - \frac{4}{m} + 2 \right|$ $A = \frac{1}{2} \left| 4 - m - \frac{4}{m} \right|$ Since we're looking for the minimum area, we can assume that $m$ will be negative (because the line passes through (1,2) and intersects both axes, meaning m must be negative to make the intercepts positive). Thus $-m$ will be positive and $4 - m - \frac{4}{m}$ will be positive. Therefore, we can remove the absolute value signs: $A = \frac{1}{2} \left( 4 - m - \frac{4}{m} \right)$

Step 4: Find the critical points by taking the derivative of A with respect to m and setting it to zero. $\frac{dA}{dm} = \frac{1}{2} \left( -1 + \frac{4}{m^2} \right)$ Setting $\frac{dA}{dm} = 0$: $-1 + \frac{4}{m^2} = 0$ $\frac{4}{m^2} = 1$ $m^2 = 4$ $m = \pm 2$ Since we deduced that m must be negative to have positive intercepts, we take the negative root, so $m = -2$.

Step 5: Verify that this value of m minimizes the area using the second derivative test. $\frac{d^2A}{dm^2} = \frac{1}{2} \left( -\frac{8}{m^3} \right)$ Evaluating at $m = -2$: $\frac{d^2A}{dm^2} = \frac{1}{2} \left( -\frac{8}{(-2)^3} \right) = \frac{1}{2} \left( -\frac{8}{-8} \right) = \frac{1}{2} > 0$ Since the second derivative is positive, the area is minimized at $m = -2$.

Step 6: Re-evaluate to check for errors. We assumed that the intercepts were positive, leading to $m=-2$. However, this gives intercepts $x=2$ and $y=4$, so the area is 4. Let's consider the general case. $A = \frac{1}{2} \left| 4 - m - \frac{4}{m} \right|$ Then $\frac{dA}{dm} = \frac{1}{2} \text{sgn}\left(4 - m - \frac{4}{m}\right) \left(-1 + \frac{4}{m^2}\right)$ Setting this to 0 gives $m = \pm 2$. If $m=2$, then $x=0$ and the area is undefined. For $m=-2$, $x=1 - \frac{2}{-2} = 2$ and $y = 2 - (-2) = 4$, so $A = \frac{1}{2} \times 2 \times 4 = 4$.

The problem statement says that the correct answer is -1/4. Let's reconsider the intercepts. If the line intersects both the positive axes, then since it passes through (1,2), $m$ must be negative. So let $m = -r$ for some $r>0$. $x = 1 - \frac{2}{m} = 1 + \frac{2}{r}$ $y = 2 - m = 2 + r$ Then $A = \frac{1}{2} (1 + \frac{2}{r}) (2 + r) = \frac{1}{2} (2 + r + \frac{4}{r} + 2) = \frac{1}{2} (4 + r + \frac{4}{r})$ To minimize $A$, we need to minimize $r + \frac{4}{r}$. By AM-GM, $\frac{r + \frac{4}{r}}{2} \ge \sqrt{r \cdot \frac{4}{r}} = 2$, so $r + \frac{4}{r} \ge 4$. Equality when $r = \frac{4}{r}$, so $r^2 = 4$ and $r = 2$ (since $r>0$). So $m = -2$. It still looks like the answer should be -2.

Step 7: Let's try another approach. The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$. Since the line passes through (1,2), $\frac{1}{a} + \frac{2}{b} = 1$. We want to minimize $A = \frac{1}{2}ab$. From $\frac{1}{a} + \frac{2}{b} = 1$, we have $\frac{1}{a} = 1 - \frac{2}{b} = \frac{b-2}{b}$, so $a = \frac{b}{b-2}$. $A = \frac{1}{2} \frac{b^2}{b-2}$. $\frac{dA}{db} = \frac{1}{2} \frac{2b(b-2) - b^2}{(b-2)^2} = \frac{1}{2} \frac{2b^2 - 4b - b^2}{(b-2)^2} = \frac{1}{2} \frac{b^2 - 4b}{(b-2)^2} = \frac{b(b-4)}{2(b-2)^2}$. Setting this to 0, we have $b=0$ or $b=4$. $b=0$ is invalid. If $b=4$, then $a = \frac{4}{4-2} = 2$. The equation is $\frac{x}{2} + \frac{y}{4} = 1$, or $2x + y = 4$. Then $y = -2x + 4$, so $m = -2$.

If the correct answer is -1/4, then the line is $y-2 = -\frac{1}{4}(x-1)$, so $4y - 8 = -x + 1$, or $x + 4y = 9$. $x$-intercept is 9, $y$-intercept is 9/4. Area is $\frac{1}{2} (9)(9/4) = \frac{81}{8} = 10.125$. When $m=-2$, the area is 4.

There might be an error in the given answer.

Common Mistakes & Tips

• Sign Errors: Be careful with signs, especially when dealing with negative slopes and intercepts.
• Absolute Values: Remember to consider the absolute value when calculating the area of the triangle, especially if you're unsure about the signs of the intercepts. However, in this case, using the assumption about the sign of $m$ allowed us to drop the absolute value signs.
• Domain Restrictions: Check for domain restrictions on the slope $m$. For example, $m$ cannot be zero, as the line would be horizontal and would not intersect the x-axis. Also, $m$ cannot be 2, as the y-intercept would be zero, implying an undefined area.

Summary

We found the x and y intercepts in terms of the slope $m$ of the line. Then, we formulated an expression for the area of the triangle formed by the origin and the intercepts. We then minimized the area by taking the derivative of the area with respect to $m$, setting it equal to zero, and solving for $m$. We verified that this value of $m$ minimizes the area using the second derivative test. The resulting slope is $m=-2$. However, this does not match the provided correct answer. After re-evaluating using an alternative approach, we continue to arrive at $m=-2$. Therefore, there is a possible error in the provided answer.

The final answer is \boxed{-2}.