Given that the straight line and the positive half axis of x-axis and y-axis intersect at two points a and B respectively, and pass through point P (5,6), the minimum area of triangle AOB and the linear equation at this time are obtained

Given that the straight line and the positive half axis of x-axis and y-axis intersect at two points a and B respectively, and pass through point P (5,6), the minimum area of triangle AOB and the linear equation at this time are obtained

The slope of the straight line K0 - 36 / k > 0 applies the mean value inequality s ≥ 1 / 2 (2 √ 25 * 36 + 60) = 60. Only when - 25K = - 36 / K, the minimum value is taken. At this time, the k = - 6 / 5 linear equation is y = - 6 (X-5) / 5 + 6 = - 6x / 5 + 12, and the general formula is 6x + 5y-60 = 0

When the area s of △ AOB (o is the origin) is the smallest, the equation of line L is solved and the minimum value of S is obtained

Let a (a, 0), B (0, b), (a, b > 0), then the equation of the line L is XA + Yb = 1, and ∵ P (4, 1) on the line L, ∵ 4A + 1b = 1 (6 points) and ∵ 1 = 4A + 1b ≥ 24ab, ∵ ab ≥ 16, ∵ s = 12ab ≥ 8, the equal sign holds if and only if 4A = 1b = 12, that is, a = 8, B = 2, the equation of ∵ line L is: x + 4y-8 = 0, smin = 8 (12 points)

Given that the image of a first-order function passes through point a (3,0) and intersects with Y-axis at point B, if the area of triangle AOB is 6, the analytic expression of the first-order function is obtained
As shown in the figure, the image of the first-order function passes through two points B (0,2) and C (- 3,0), the line x = 4 intersects the image of the first-order function at point a, and intersects with the X axis at point m (1) to find the analytic expression of the first-order function (2) to find the area of the trapezoid ABOM

Let the analytic expression of a function be y = KX + B. the intersection of the graph of this function and the Y axis B (0, b). S △ AOB = (1 / 2) oaob = 6. (1 / 2) * 3 * b = 6. B = 4. By substituting a (3,0) into the equation of y = KX + 4, we can get: 3K + 4 = 0. K = - 4 / 3