Through point (2,1), make a straight line a and X axis, the positive half axis of Y axis intersects at two points a and B, and calculate the minimum area of triangle AOB

Through point (2,1), make a straight line a and X axis, the positive half axis of Y axis intersects at two points a and B, and calculate the minimum area of triangle AOB

Let the linear equation be x / A + Y / b = 1
Then 2 / A + 1 / b = 1
Area AB / 2 = AB / 2 * (2 / A + 1 / b) = (a + 2b) / 2 = (a + 2b) (2 / A + 1 / b) / 2
=(4 + A / B + 4B / a) / 2 > = 4 (inequality theorem)

When the area of △ ABC is the smallest, the equation of line L is obtained

Let a (a, 0) B & nbsp; (0, b) (a, b > 0), then the equation of the line L is XA + Yb = 1, from the meaning of the problem, the point is on the line, so 1A + 2B = 1, from the basic inequality, we get 1 = 1A + 2B ≥ 22ab, ｜ ab ≥ 8, then s △ AOB = 12ab ≥ 4 & nbsp; if and only if & nbsp; When 1A = 2B, i.e. a = 2, B = 4, take "=", therefore, when the area of △ AOB is the smallest, the equation of line L is x2 + Y4 = 1, i.e. 2x + y-4 = 0

Given that the line L passes through P (- 1, - 2) and intersects with the negative half axis of x-axis and y-axis at two points a and B, the minimum area of △ AOB is obtained, and the equation of the line L is obtained

Let y + 2 = K (x + 1)
Let y = 0, then x = 2 / k-1, OA length is 1-2 / K
Let x = 0, y = K-2, OB length be 2-k
Triangle area s = 0.5 * (1-2 / k) * (2-k), and then give you the basic inequality