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 triangle OPQ is least, then the slope of the line PQ is

Options

A. `-(1)/(4)`
B. -4
C. -2
D. `-(1)/(2)`

Explanation

To solve the problem step by step, we will find the slope of the line that minimizes the area of triangle OPQ formed by the line through point (1, 2) and the coordinate axes.

### Step 1: Set up the equation of the line
The line passes through the point (1, 2) and has a slope \( m \). The equation of the line can be expressed using the point-slope form:
\[ y - 2 = m(x - 1) \]
Rearranging this gives:
\[ y = mx - m + 2 \]

### Step 2: Find the x-intercept and y-intercept
To find the intercepts, we set \( y = 0 \) for the x-intercept (point Q) and \( x = 0 \) for the y-intercept (point P).

- **For the x-intercept (Q)**:
\[ 0 = mx - m + 2 \implies mx = m - 2 \implies x = \frac{m - 2}{m} \quad (m \neq 0) \]
Thus, the coordinates of point Q are \( \left(\frac{m - 2}{m}, 0\right) \).

- **For the y-intercept (P)**:
\[ y = m(0) - m + 2 \implies y = 2 - m \]
Thus, the coordinates of point P are \( (0, 2 - m) \).

### Step 3: Calculate the area of triangle OPQ
The area \( A \) of triangle OPQ can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is the x-intercept \( \frac{m - 2}{m} \) and the height is the y-intercept \( 2 - m \):
\[ A = \frac{1}{2} \times \frac{m - 2}{m} \times (2 - m) \]
Simplifying this gives:
\[ A = \frac{(m - 2)(2 - m)}{2m} \]
Expanding the numerator:
\[ A = \frac{-(m^2 - 4)}{2m} = \frac{4 - m^2}{2m} \]

### Step 4: Minimize the area
To find the minimum area, we differentiate \( A \) with respect to \( m \) and set the derivative to zero:
\[ \frac{dA}{dm} = \frac{d}{dm}\left(\frac{4 - m^2}{2m}\right) \]
Using the quotient rule:
\[ \frac{dA}{dm} = \frac{(2m)(-2m) - (4 - m^2)(2)}{(2m)^2} \]
Setting the numerator equal to zero:
\[ -4m^2 + 8 - 2m^2 = 0 \implies -6m^2 + 8 = 0 \implies 6m^2 = 8 \implies m^2 = \frac{4}{3} \]
Thus, \( m = \pm \sqrt{\frac{4}{3}} \).

### Step 5: Determine the slope
Since we are looking for the slope that minimizes the area, we find:
\[ m = \pm 2 \]
Given that the area is minimized, we choose the negative slope (as the triangle is formed in the first quadrant):
\[ m = -2 \]

### Final Answer
The slope of the line PQ that minimizes the area of triangle OPQ is:
\[ \boxed{-2} \]