Question

一道初三数学题。。。求第二小问过程答案QAQ

Answer 1

y=kx-3
令y=0得,x=3/k
S=1/2lOAl*lOBl=3
1/2*3/k*3=3
k=3/2,由于k<0
k=-3/2
y=-3/2x-3
2)C(x1,12/x1),D(x2,12/x2),
由于,四边形ABCD是平行四边形
所以,AB//CD,且AB=CD
Kcd=Kab=-3/2
Kcd=(12/x1-12/x2)/(x1-x2)
(12/x1-12/x2)/(x1-x2)=-3/2
整理得:
x1x2=8.....................................(1)
lOBl/lOAl=3/2.lOBl=3
lOAl=2
AB^2=OA^2+OB^2=13
CD=AB
CD^2=AB^2=13
因CD^2=(x1-x2)^2+(12/x1-12/x2)^2=13
(x1-x2)^2+144(x1-x2)^2/(x1x2)^2=13
(x1-x2)^2+144(x1-x2)^2/(x1x2)^2=13..............(2)
把(1)代入(2)得
(x1-x2)^2+144(x1-x2)^2/64=13
(x1-x2)^2+9(x1-x2)^2/4=13
(x1-x2)^2=4
(x1-x2)^2+4x1x2=4+4x1x2
(x1+x2)^2=4+4x1x2=4+4*8=36
(x1+x2)^2=36
x1+x2=6 或 x1+x2=-6
解,x1+x2=6和x1x2=8得
x1=2,x2=4或x1=4,x2=2
解,x1+x2=-6和x1x2=8得
x1=-2,x2=-4或x1=-4,x2=-2
所以,C(2,6),D(4,3) 或C(4,3),D(2,6)
或,C(-2,-6),D(-4,-3) 或C(-4,-3),D(-2,-6)