(Note: in the following answers, 1, M 1, x 1, y 1, A2, M2, N2, x2, y2 "1and 2 represent A, M, N and Y2 respectively.
Case 1: vector ba1/vector CD, angle A= angle D=90 degrees, let A 1(M 1, N 1).
Vector BA 1= vector OA 1- vector OB=(M 1, n 1)-(6, 1) = (m 1-6, n1.
Vector CD= vector OD- vector OC = (2 2,5)-(3,3) = (-1,2)
Vector DA 1= vector OA 1- vector OD=(M 1, n1)-(2,5) = (m1-2, N 1-5).
Because: vector BA 1// vector CD.
Therefore, x1y2-x2y1= 0,2m1-12-1+n1= 0 [1].
Because: the vector DA 1 is perpendicular to the vector CD.
So: from vector DA 1* vector CD = 0,2-m1+2n1-10 = 0 [2].
[1][2] is simultaneous, and the solution is: M 1=3.6, N 1=5.8.
Case 2: vector A2D// vector BC, angle A2= angle B=90 degrees, and let A2(M2, N2) at this time.
Vector A2D= vector OD- vector OA2 = (2 2,5)-(m2,N2 N2) = (2-m2-m2,5-N2)
Vector BC= vector OC- vector OB = (3 3,3)-(6,1) = (-3,2)
Vector BA2= vector OA2- vector OB=(M2, N2)-(6 1)=(m2-6, N2- 1)
Because: vector A2D// vector BC
So: x1y2-x2y1= 0,4-2m2+15-3N2 = 0 [3].
Because the vector BA2 is perpendicular to the vector BC.
Therefore, from vector BA2* vector BC=0, -3M2+ 18+2N2-2=0 [4].
[3][4] Simultaneous, the solution is: 03 of m2 = 65438+86, and 25 of N2 = 13.
Therefore, when M 1=3.6, N 1=5.8 or m2 = 03 of 65438+86, N2 = 13 of 25, the quadrilateral ABCD is a right trapezoid.