Q. Origin containing angle bisector of two lines $L_{1} \equiv a_{1}x + b_{1}y + c_{1} = 0$ and $L_{2} \equiv a_{2}x + b_{2}y + c_{2} = 0$ (where $c_{1}c_{2} < 0$) is
A) $\frac{a_{1}x + b_{1}y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}} = \frac{a_{2}x + b_{2}y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}$
B) $\frac{a_{1}x + b_{1}y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}} = - \frac{a_{2}x + b_{2}y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}$
C) $\frac{a_{1}x + b_{1}y + c_{1}}{a_{1}^{2} + b_{1}^{2}} = \frac{a_{2}x + b_{2}y + c_{2}}{a_{2}^{2} + b_{2}^{2}}$
D) Depends on the value of $c_{1}$ and $c_{2}$
Explanation
The correct answer is Option B: $\frac{a_{1}x + b_{1}y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}} = - \frac{a_{2}x + b_{2}y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}$.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.