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Line through \mathrm{P}(\mathrm{a}, \mathrm{b}) is equally inclined to lines 2 \mathrm{x}+\mathrm{y}+1=0 and \mathrm{x}+2 \mathrm{y}+3=0 and \mathrm{(x, y)} is on the line and \mathrm{b-a=\lambda_{1}, b+a=\lambda_{2}} , then \mathrm{\lambda_{1} \, \&\, \lambda_{2}} respectively equals

Option: 1

\mathrm{y+x, y-x}


Option: 2

\mathrm{y, x}


Option: 3

\mathrm{y-x, y+x}


Option: 4

\mathrm{y},-\mathrm{x}


Answers (1)

best_answer

Required lines are parallel to the angle bisector

\frac{2 \mathrm{x}+\mathrm{y}+1}{\sqrt{5}}= \pm \frac{\mathrm{x}+2 \mathrm{t}+3}{\sqrt{5}}

Slopes of bisectors are \pm 1,
\mathrm{y-b=1(x-a) \Rightarrow y+x=b+x}
\mathrm{\Rightarrow \quad (b-a, b+a) \Rightarrow(y-x, y+x) }
\mathrm{\therefore }   (C)  is correct answer.

Posted by

Divya Prakash Singh

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