推 brian821117 :辛苦了感恩~~~ 12/28 00:37
※ 引述《brian821117 (chaozoo)》之銘言：
: 若兩直線
: L1:mX+3Y-2=0與L2:3X+mY+1=0
: 相交於一點P，滿足P點在第三象限，則整數m值共有幾個?
: Ans:2
: 求解拜託了
mx+3y=2 ----(1)
3x+my=-1 ----(2)
2m+3
(1)·m-(2)·3 => (m^2 - 9)x=2m+3, x= ─────
m^2 -9
m+6
(1)·3-(2)·m => (9 - m^2)y=6+m => y= ─────
9-m^2
x<0 => (2m+3)(m^2-9)<0 => (2m+3)(m+3)(m-3)<0 => m<-3 or -1.5<m<3
y<0 => (m+6)(9-m^2)<0 => (m+6)(m+3)(m-3)>0 => -6<m<-3 or m>3
由上兩式 聯立可得 -6<m<-3
又 ∵m為整數, m=-5 or -4
Ans: 2個
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