(1)为书写明了,假定AB与y轴的角点为H;
因 BC'=BC=5,HB=OC=4,∴HC'=√(BC'²-HB²)=√(5²-4²)=3,则 OC'=CB+HC'=5+3=8;
直线C'D的斜率 k=OC/DO=8/6,∴C'D的解析式:y=4(x+6)/3,即 y=4x/3+8;
(2)运动时间为t时,DE=5t,(0
(1)为书写明了,假定AB与y轴的角点为H;
因 BC'=BC=5,HB=OC=4,∴HC'=√(BC'²-HB²)=√(5²-4²)=3,则 OC'=CB+HC'=5+3=8;
直线C'D的斜率 k=OC/DO=8/6,∴C'D的解析式:y=4(x+6)/3,即 y=4x/3+8;
(2)运动时间为t时,DE=5t,(0