question_answer

The lines y\[y={{m}_{1}}x,\] \[y={{m}_{2}}x,\]and \[y={{m}_{3}}x,\]make equal intercepts on the line\[x+y=1\]. Then

A)  \[2(1+{{m}_{1}})(1+{{m}_{3}})=(1+{{m}_{2}})(2+{{m}_{1}}+{{m}_{3}})\]

B)  \[(1+{{m}_{1}})(1+{{m}_{3}})=(1+{{m}_{2}})(1+{{m}_{1}}+{{m}_{3}})\]

C)  \[(1+{{m}_{1}})(1+{{m}_{2}})=(1+{{m}_{3}})(2+{{m}_{1}}+{{m}_{3}})\]

D)  \[2(1+{{m}_{1}})(1+{{m}_{3}})=(1+{{m}_{2}})(1+{{m}_{1}}+{{m}_{3}})\]

Correct Answer: A

Solution :

[a] Solving the equations of the liners, we get  \[A\equiv \left( \frac{1}{1+{{m}_{1}}},\frac{{{m}_{1}}}{1+{{m}_{1}}} \right),\] \[\,C\equiv \left( \frac{1}{1+{{m}_{3}}},\frac{{{m}_{3}}}{1+{{m}_{3}}} \right)\] If AB=BC. Then the med pint of AC lies on \[y={{m}_{2}}x\]