Question

If $\mathrm y=\sqrt3\mathrm x+1$ is an angle bisector of $L_{1}$ and $L_{2}$ and $L_{1}$ is $\mathrm y=\dfrac{\mathrm x}{\sqrt3}+1$, then $L_{2}$ is-
A. $x=0$
B. $y=0$
C. $x=1$
D. $y=1$

Answer 1

Hint: The equation of the angle bisector of two given lines is obtained by the formula-
$\dfrac{{\mathrm a}_1\mathrm x+{\mathrm b}_1\mathrm y+{\mathrm c}_1}{\sqrt{{\mathrm a}_1^2+{\mathrm b}_1^2}}=\pm\left(\dfrac{{\mathrm a}_2\mathrm x+{\mathrm b}_2\mathrm y+{\mathrm c}_2}{\sqrt{{\mathrm a}_2^2+{\mathrm b}_2^2}}\right)$ Also, the slope of a line with angle $p$ is given by $tanp$.

Complete step by step answer:
 

The angle bisector of line is given by-
$\mathrm y=\sqrt3\mathrm x+1$
By comparing with $y = mx + c$
So the slope of the line is $\tan^{-1}\sqrt3=60^{\mathrm o}$
The equation of line $L_{1}$ is given by-
$\mathrm y=\dfrac{\mathrm x}{\sqrt3}+1$
So the slope of the line is $\tan^{-1}\left(\dfrac1{\sqrt3}\right)=30^{\mathrm o}$
From the figure we can clearly see that $L_{1}$ is at an angle $30^{o}$ and the angle bisector is at an angle $60^{o}$. Hence, $L_{2}$ will be at an angle $90^{o}$. In the figure, it is visible that $L_{2}$ lies on the y-axis. Hence the equation of $L_{2}$ is x = 0.
Note: We can also solve this problem using the formula of angle bisector but it is a very lengthy method. Instead, by using the diagram and geometry, we can directly find the equation of $L_{2}$.