Question

${\theta }_1$ and ${\theta }_2$ are the inclination of lines $L_1$ and $L_2$ with x-axis. If $L_1$ and $L_2$ pass through $P(x_1,y_1)$, then equation of one of the angle bisector of these lines is

• A. $\frac{x-x_1}{\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}=\frac{y-y_1}{\sin \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}$
• B. $\frac{x-x_1}{-\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}=\frac{y-y_1}{\sin \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}$
• C. $\frac{x-x_1}{\sin \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}=\frac{y-y_1}{\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}$
• D. $\frac{x-x_1}{-\sin \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}=\frac{y-y_1}{\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}$

Answer 1

Answer: (A), (D)

The correct options are
A $\frac{x-x_1}{\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}=\frac{y-y_1}{\sin \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}$
D $\frac{x-x_1}{-\sin \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}=\frac{y-y_1}{\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}$

Angle bisectors will make angle $\frac{{\theta }_1+{\theta }_2}{2}$ or $(\frac{\pi }{2}+\frac{{\theta }_1+{\theta }_2}{2})$
hence the equation will be either : $\frac{x-x_1}{\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}=\frac{y-y_1}{\sin \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}$
or : $\frac{x-x_1}{-\sin \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}=\frac{y-y_1}{\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}$