As shown in the figure, it is known that the quadrilateral ABCD is a parallelogram, BC = 2Ab. The coordinates of two points a and B are (- 1,0), (0,2), respectively, and the coordinates of two points c and D are on the image of inverse scale function y = KX (k < 0), then K is equal to___ ．

As shown in the figure, it is known that the quadrilateral ABCD is a parallelogram, BC = 2Ab. The coordinates of two points a and B are (- 1,0), (0,2), respectively, and the coordinates of two points c and D are on the image of inverse scale function y = KX (k < 0), then K is equal to___ ．

Let the coordinates of point C be (a, KA), (K ＜ 0), the coordinates of point d be (x, y), ∵ the quadrilateral ABCD is a parallelogram, the midpoint coordinates of AC and BD are the same, ∵ (A-12, k2a) = (X2, y + 22), then x = A-1, y = k-2aa, substituting y = KX, we can get k = 2a-2a2 ① In RT △ AOB, ab = oa2 + ob2 = 5, ∵ BC = 2Ab = 25, so BC2 = (0-A) 2 + (Ka-2) 2 = (25) 2, we can get: A4 + k2-4ka = 16a2, we can get: A2 = 4, ∵ a < 0, ∵ a = - 2, ∵ k = - 4-8 = - 12 by substituting ① k = 2a-2a2, so the answer is: - 12 Let the coordinates of point C be (- A, 2 + b), and the coordinates of point d be (- 1-A, b), (a > 0, b > 0) according to the geometric meaning of K, | - a ×| 2 + B | = | - 1-A ×| B |, it is sorted out that 2A + AB = B + AB, and the solution is b = 2A |=The graph of the function is in the second quadrant, so k = - 12