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The equation of the bisector of angle between the lines \mathrm{x+y=1} and \mathrm{7 x-y=3}  that contains the point \mathrm{(2,3)} is 

Option: 1

\mathrm{3 x+y-1=0}


Option: 2

\mathrm{x-3 y+1=0}


Option: 3

\mathrm{2 x+y-1=0}


Option: 4

\mathrm{x-2 y+1=0}


Answers (1)

best_answer

Equation of bisector

\mathrm{\left|\frac{x+y-1}{\sqrt{2}}\right|=\left|\frac{7 x-y-3}{\sqrt{50}}\right|}

Now at \mathrm{(2,3) x+y-1>0} and \mathrm{7 x-y-3>0}.

Hence the equation of bisector contains the point \mathrm{(2,3)} is \mathrm{\frac{x+y-1}{\sqrt{2}}=\frac{7 x-y-3}{\sqrt{50}}}
 

Hence (B) is the correct answer.

 

Posted by

sudhir.kumar

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