解

(1)如图,设D(x,y),

　　则由CD⊥AB,BC∥AD,可知{■(k_CD "·" k_AB="-" 1"," @k_CB=k_AD "," )┤

　　得{■(y/(x"-" 3) "·" (2+1)/(2"-" 1)="-" 1"," @(2"-" 0)/(2"-" 3)=(y+1)/(x"-" 1) "," )┤

　　解得{■(x=0"," @y=1"," )┤即点D坐标为(0,1).

　　(2)∵kAC=(0"-(-" 1")" )/(3"-" 1)=1/2,kBD=(2"-" 1)/(2"-" 0)=1/2,∴kAC=kBD.

　　∴AC∥BD,∴四边形ACBD为平行四边形.

　　而kBC=(2"-" 0)/(2"-" 3)=-2,∴kBC·kAC=-1.

　　∴AC⊥BC,∴四边形ACBD是矩形.

　　∵DC⊥AB,∴四边形ACBD是正方形.

B组　能力提升

1.若过点A(-2,2),B(5,0)的直线与过点P(2m,1),Q(-1,m)的直线平行,则m的值为(　　)

A.-1 B.3 C.2 D.1/2

解析由已知kAB=kPQ,得2/("-" 2"-" 5)=(1"-" m)/(2m+1),

　　解得m=3,故选B.

答案B

2.已知直线l1:mx+4y-2=0与l2:2x-5y+n=0互相垂直且垂足为(1,p),则m-n+p的值为(　　)

A.24 B.20 C.0 D.-8

解析因为l1⊥l2,所以2m+4×(-5)=0,解得m=10,又点(1,p)在l1上,所以10+4p-2=0,即p=-2,因为点(1,p)在l2上,所以2×1-5p+n=0,得n=-12.

　　所以m-n+p=10-(-12)+(-2)=20.

答案B

3.已知点O(0,0),A(0,b),B(a,a3).若△OAB为直角三角形,则必有(　　)