数学题
已知直线l1、l2的方程分别l1:A1x+B1y+C=0,l2:A2x+B2y+C=0,且A1B1C1≠0,A2B2C2≠0,A1B2-A2B1=0,B1C2-B2C1≠0.求证l1∥l2.
直线L1: A1X+B1Y+C1=0===>Y=-A1X/B1-C1 斜率K1=-A1/B1 直线L2: A2X+B2Y+C2=0===>Y=-A2X/B2-C2 斜率K2=-A2/B2 ∴K1×K2=A1A2/B1B2 ∵A1B2-A2B1=0===>A1B2=A2B1 ∴K1×K2=(A2/B2)²=(A1/B1)² ∴K1=K2 ∵A1B1C1≠0,∴A1≠0,B1≠0,C1≠0(直线不过原点) ∵A2B2C2≠0,∴A2≠0,B2≠0,C2≠0(直线不过原点) ∵B1C2-B2C1≠0,∴B1C2≠B2C1 ∴直线L1≠直线L2(亦即两直线不重合) ∴L1∥L2(斜率相等的两条直线平行)
答:1.因为A1B1C1≠0,A2B2C2≠0,所以B1≠0,B2≠0,B1B2≠0 因为A1B2-A2B1=0,B1C2-B2C1≠0,所以(A1B2/B1B2)...详情>>
答:详情>>