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It is known that point (1,3) is on the image of function y = KX (x > 0), the edge BC of rectangle ABCD is on the X axis, e is the midpoint of diagonal BD, the image of function y = KX (x > 0) passes through two points a and E, and the abscissa of point E is m. The following problems are solved: (1) find the value of K; (2) find the abscissa of point C; (3) find the value of m when ∠ abd = 45 °
(1) The coordinate of point (1,3) can be substituted into y = KX by the function y = KX image passing through point (1,3) to get k = 3; (2) connect AC, then AC passes through e, eg ⊥ BC intersects BC with point G ∵ the abscissa of point E is m, e is on hyperbola y = 3x, the ordinate of E is y = 3M, ∵ e is the midpoint of BD, ≁ e is the midpoint of AC, ≁ BG = GC = 12bc, ≁ AB = 2EG = 6m, that is, the ordinate of point a is 6m The abscissa of a is 12M, OB = 12M, i.e. BG = GC = m-12m = 12M, CO = 12m + M = 32m, C (32m, 0). (3) when ∠ abd = 45 ° and ab = ad, there is 6m = m, i.e. M2 = 6, the solution M1 = 6, M2 = - 6 (rounding off), M = 6
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