Question

两条直线L1:(m+3)x+2y=5-3m,L2:4x+(5+m)y=16，则m为何值时，

Answer 1

L1:(m+3)x+2y=5-3m,那么斜率k1=-(m+3)/2
L2:4x+(5+m)y=16,那么斜率k2=-4/(5+m)
l1到l2
的角是45度,则有tan45=(k2-k1)/(1+k1k2)=(-4/(5+m)+(m+3)/2)/(1+4/(5+m)*(m+3)/2)=1
即有:-8+(m+3)(m+5)=2(5+m)+4(m+3)
-8+m^2+8m+15=10+2m+4m+12
m^2+2m-15=0
(m+5)(m-3)=0
m+5不=0,则有m=3
当m+5=0
时,l2:x=4,L1:-2x+2y=20,即有y=x+10,二直线成45度的角,符合
即有m=-5或m=3
(2)L1与L2的夹角是45度,则有tan45=|(k1-k2)/(1+k1k2)|
即有|(-4/(5+m)+(m+3)/2)/(1+4/(5+m)*(m+3)/2)|=1
即有|-8+(m+3)(m+5)|=|2(5+m)+4(m+3)|
|m^2+8m+7|=|6m+22|
m^2+8m+7=6m+22,或m^2+8m+7=-6m-22
(m+5)*(m-3)=0
或m^2+14m+29=0
即有m=-5或3,
(m+7)^2=20
m=-7土2根号5
即有m=-5或3或-7土2根号5