Given that the point C is the inverse scale function, the perpendicular foot of the vertical line drawn from the point C to the coordinate axis is ab, and the area of the quadrilateral aobc is 6, the analytic solution of the inverse scale function is obtained

Given that the point C is the inverse scale function, the perpendicular foot of the vertical line drawn from the point C to the coordinate axis is ab, and the area of the quadrilateral aobc is 6, the analytic solution of the inverse scale function is obtained

Let: the analytic expression of the inverse proportion function be y = K / x, C (x, y) be a point on the hyperbola, the perpendicular of the x-axis intersects the x-axis at a through C, the perpendicular of the y-axis intersects the y-axis at B, then the quadrilateral aobc is a rectangle, Ao = BC = | x |, CA = Bo = | y |, so the area of the quadrilateral aobc is s = | x | y |, because s = 6, that is, xy = 6, or xy = - 6, so the inverse proportion

As shown in the figure, given that point C is a point on the inverse scale function y = - 6x, a vertical line is drawn through point C to the coordinate axis, and the vertical feet are a and B respectively, then the area of the quadrilateral aobc is______ ．

Since the point C is a point on the inverse scale function y = - 6x, the area of the quadrilateral aobc s = | K | = 6

As shown in the figure, y = KX is known. If the figure intersects with the image of y = 4 / X at two points a and B, and AC is perpendicular to the X axis, then s triangle ABC =?

If two graphs intersect, we know that K is greater than 0. If two equations are simultaneous, we can solve x = root (4 / k) = 2 / root K, y = 4 / x = 2 * root K. The combination of these two roots is the coordinates of the point in the first quadrant
Because y = KX passes through the origin, points a and B are symmetric about the origin, so s triangle ABC = XY / 2 + (- x) * (- y) / 2 = xy = 1

It is known that the three vertices of triangle ABC are all on the image of inverse scale function y = 1 / x, and it is proved that its perpendicular center h is also on the image of this function

∵ a, B and C are all on y = 1 / x, let a, B and C coordinate in turn: (a, 1 / a), (B, 1 / b), (C, 1 /). Let h coordinate be (x, y). It is easy to get: the slope of AB = (1 / A-1 / b) / (a-b) = - 1 / (AB), the slope of BC = (1 / B-1 / C) / (B-C) = - 1 / (BC), ah