已知l1:(3+m)x+4y=5-3m,l2:2x+(5+m)y=8,求m为何值l1与l2相交〔5-3M -(3+M)X〕/4=(8-2x)/(5+m) 不晓得这步k1=-(3+m)/4,k2=-2/(5+m) 怎么得到的
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![已知l1:(3+m)x+4y=5-3m,l2:2x+(5+m)y=8,求m为何值l1与l2相交〔5-3M -(3+M)X〕/4=(8-2x)/(5+m) 不晓得这步k1=-(3+m)/4,k2=-2/(5+m) 怎么得到的](/uploads/image/z/10405959-15-9.jpg?t=%E5%B7%B2%E7%9F%A5l1%EF%BC%9A%EF%BC%883%2Bm%29x%2B4y%3D5-3m%2Cl2%EF%BC%9A2x%2B%285%2Bm%29y%3D8%2C%E6%B1%82m%E4%B8%BA%E4%BD%95%E5%80%BCl1%E4%B8%8El2%E7%9B%B8%E4%BA%A4%E3%80%945-3M+-%283%2BM%29X%E3%80%95%2F4%3D%288-2x%29%2F%285%2Bm%29+%E4%B8%8D%E6%99%93%E5%BE%97%E8%BF%99%E6%AD%A5k1%3D-%283%2Bm%29%2F4%2Ck2%3D-2%2F%285%2Bm%29+%E6%80%8E%E4%B9%88%E5%BE%97%E5%88%B0%E7%9A%84)
已知l1:(3+m)x+4y=5-3m,l2:2x+(5+m)y=8,求m为何值l1与l2相交〔5-3M -(3+M)X〕/4=(8-2x)/(5+m) 不晓得这步k1=-(3+m)/4,k2=-2/(5+m) 怎么得到的
已知l1:(3+m)x+4y=5-3m,l2:2x+(5+m)y=8,求m为何值l1与l2相交
〔5-3M -(3+M)X〕/4=(8-2x)/(5+m)
不晓得这步
k1=-(3+m)/4,k2=-2/(5+m) 怎么得到的
已知l1:(3+m)x+4y=5-3m,l2:2x+(5+m)y=8,求m为何值l1与l2相交〔5-3M -(3+M)X〕/4=(8-2x)/(5+m) 不晓得这步k1=-(3+m)/4,k2=-2/(5+m) 怎么得到的
直线相交就是斜率不相等或者一条斜率不存在而另一条斜率存在
l1:(3+m)x+4y=5-3m,l2:2x+(5+m)y=8
所以k1=-(3+m)/4,k2=-2/(5+m)
若相等
-(3+m)/4=-2/(5+m)
(m+3)(m+5)=8
m^2+8m+7=0
m=-1,m=-7
所以相交就是m≠-1,m≠-7
斜率不存在就是垂直x轴,即y的系数是0
l1不可能
l2则m=-5,符合m≠-1,m≠-7
所以m≠-1,m≠-7
当m=-3时 l1与l2相交 当m不等与-3时 两直线平衡条件为4/(3+m)=(5+m)/2 得m=-1或m=-7 所以当m不等与-1且不等于-7时 l1与l2相交
L1:(3+M)X+4Y=5-3M ==>y=〔5-3M -(3+M)X〕/4
L2:2X+(5+M)Y=8 ==>y=(8-2x)/(5+m)
相交
〔5-3M -(3+M)X〕/4=(8-2x)/(5+m)
得m≠-1,-7