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, the slope of the line PQ is

Equation of line passing through (1,2) with slope m is y - 2 = m(x - 1).
Let the line intersect the x-axis at P(x, 0) and the y-axis at Q(0, y).
Then, -2 = -m, so m = 2. The y-intercept is y = m(0 - 1) + 2 = 2 - m
At P, 0 - 2 = m(x - 1), so x = 1 - 2/m
At Q, y - 2 = m(0 - 1), so y = 2 - m
Area of triangle OPQ = (1/2)xy = (1/2)(1 - 2/m)(2 - m)
Let A = (1/2)(2 - m - 4/m + 2) = (1/2)(4 - m - 4/m)
A' = (1/2)(-1 + 4/m^2) = 0
4/m^2 = 1
m^2 = 4
m = ±2
If m = 2, A = 0. This is a degenerate triangle.
If m = -2, x = 2, y = 4, A = 4
For minimum area, we consider the distance from (1,2) to the line x+y=0.
Let the line be y - 2 = m(x - 1)
Then y = mx - m + 2
The intercepts are (m-2)/m and 2-m
Area = (1/2) | (m-2)/m * (2-m) | = (1/2) | (m-2)^2 / m |
Let f(m) = (m-2)^2 / m = m + 4/m - 4
f'(m) = 1 - 4/m^2 = 0
m^2 = 4
m = ±2
f''(m) = 8/m^3
f''(2) > 0, so m = 2 is a minimum.
But the area is 0 at m = 2
f''(-2) < 0, so m = -2 is a maximum.
Thus the slope is -2.