1).令x=0得y=1,因此A点的坐标为(0,1);x=3时y=-45/4+51/4+1=10/4=5/2,故B点的坐标
为(3,5/2);KAB=(5/2-1)/(3-0)=1/2,故AB所在直线的方程为y=(1/2)x+1.
2).S=-(5/4)t²+(17/4)t+1-[(1/2)a+1]=-(5/4)t²+(15/4)t=(5/4)(3-t)t,0≦t≦3.
3).要使BCMN为平行四边形只须CM∥BN,即只须KCM=KBN;KCM=(t/2+1)/(t-3);
KBN=(-5t²/4+17t/4+1-5/2)/(t-3)=(-5t²/4+17t/4-3/2)/(t-3);故得等式:
t/2+1=-5t²/4+17t/4-3/2,5t²/4-15t/4+5/2=0,即有t²-3t+2=(t-2)(t-1)=0,故得t₁=1;t₂=2;
即当t=1秒或2秒时BCMN是平行四边形.
t=1时,︱MN︱=-5/4+17/4+1-(1/2+1)=5/2,︱CM︱=√[(3-1)²+(3/2)²]=√(25/4)=5/2;
故︱MN︱=︱CM︱,所以BCMN是菱形.
t=2时,︱MN︱=-5+17/2+1-3/2=3,︱CM︱=√[(3-2)²+2²]=√5;
故︱MN︱≠︱CM︱,所以BCMN不是菱形.