求过l1:3x+5y-10=0和l2:x+y+1=0的交点
则有:
3x+5y-10=0``````````1
x+y+1=0``````````````2
联立1、2解得:
x=-7.5,y=6.5
且平行于l3:x+2y-5=0 则有:k=-0.5
所以此直线方程为:
y=-0.5(x+7.5)+6.5
即:y=-0.5x+5
求过l1:3x+5y-10=0和l2:x+y+1=0的交点
则有:
3x+5y-10=0``````````1
x+y+1=0``````````````2
联立1、2解得:
x=-7.5,y=6.5
且平行于l3:x+2y-5=0 则有:k=-0.5
所以此直线方程为:
y=-0.5(x+7.5)+6.5
即:y=-0.5x+5