A) \[x-y+1=0\]
B) \[x+y-1=0\]
C) \[x-y-1=0\]
D) \[x+y+1=0\]
Correct Answer: C
Solution :
| Since, \[y=mx+\frac{1}{m}\] or \[{{m}^{2}}h-mk+1=0,\] |
| We have \[{{m}_{1}}+{{m}_{2}}=\frac{k}{h},\] and \[{{m}_{1}}{{m}_{2}}=\frac{1}{h}\] |
| Given \[{{\theta }_{1}},{{\theta }_{2}}=\frac{\pi }{4}\] |
| \[\therefore \tan ({{\theta }_{1}}+{{\theta }_{2}})=\tan \frac{\pi }{4}\] |
| \[\Rightarrow \frac{{{m}_{1}}+{{m}_{2}}}{1-{{m}_{1}}{{m}_{2}}}=1\]\[\Rightarrow \frac{k}{h}=1-\frac{1}{h}\] |
| \[\Rightarrow y=x-1\]\[\Rightarrow x-y-1=0\] |
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