问题
解答题
四边形ABCD中,
(1)若
(2)满足(1)的同时又有
|
答案
=(x,y)BC
=-DA
=-(AD
+AB
+BC
)=-(x+4,y-2)=(-x-4,-y+2)CD
(1)∵
∥BC DA
∴x•(-y+2)-y•(-x-4)=0,
化简得:x+2y=0;
(2)
=AC
+AB
=(x+6,y+1),BC
=BD
+BC
=(x-2,y-3)CD
∵
⊥AC BD
∴(x+6)•(x-2)+(y+1)•(y-3)=0
化简有:x2+y2+4x-2y-15=0,
联立x+2y=0 x2+y2+4x-2y-15=0
解得
或x=-6 y=3 x=2 y=-1
∵
∥BC DA
⊥AC BD
则四边形ABCD为对角线互相垂直的梯形
当x=-6 y=3
=(0,4) AC
=(-8,0)BD
此时SABCD=
•|1 2
|•|AC
|=16BD
当x=2 y=-1
=(8,0) AC
=(0,-4),BD
此时SABCD=
•|1 2
|•|AC
|=16.BD