直线(1+2m)x+(1-m)y-3-3m=0恒过哪一点?请写下解法

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直线(1+2m)x+(1-m)y-3-3m=0恒过哪一点?请写下解法
直线(1+2m)x+(1-m)y-3-3m=0恒过哪一点?
请写下解法

直线(1+2m)x+(1-m)y-3-3m=0恒过哪一点?请写下解法
(1+2m)x+(1-m)y-3-3m=0
x+2mx+y-my-3-3m=0
x+y+(2x-y-3)m-3=0
恒过定点,则与m的取值无关,所以可得含m项的系数为0,可得:
2x-y-3=0 ····················1
此时原方程可化为:
x+y-3=0·····················2
联立1、2两式得:
x=2,y=1
综上可得直线恒过定点(2,1).

x+y-3+m(2x-y-3)=0
x+y-3=0
2x-y-3=0
∴x=2
y=1
恒通过(2,1)

(1+2m)x+(1-m)y-3-3m=0
x+2mx+y-my-3-3m=0
(2x-y-3)m+(x+y-3)=0
则:
2x-y-3=0且x+y-3=0
解得:
x=2、y=1
即过定点(2,1)

(1+2m)x+(1-m)y-3-3m=0
x+2mx+y-my-3-3m=0
所以x+y=3且2x-y=3
即x=2 y=1
所以恒过(2,1)

化为m(2x-y-3)+(x+y-3)=0
2x-y-3=0 x+y-3=0
解得x=2 y=1
即恒过(2,1)这个点