Question

Sol. [Key Concept: The two angle bisectors of lines with slopes $m,m'$ through $(a,b)$ have slopes $\frac{m + m'}{1 - mm'}$ and $\frac{mm' - 1}{m + m'}$.]

$$m = tan\theta_{1},m^{'} = tan\theta_{2}$$

inclination of given lines are $2\theta_{1}$ and $2\theta_{2}$

∴ inclination of angle bisectors $\theta_{1} + \theta_{2}$ or

$$\theta_{1} + \theta_{2} + \frac{\pi}{2}$$

$\ $ slopes will be $\frac{m + m^{'}}{1 - mm^{'}},\frac{mm^{'} - 1}{m + m^{'}}$

Eqn. of bisectors are

$$\begin{matrix} & y - b = \frac{\left( m + m^{'} \right)}{\left( 1 - mm^{'} \right)}(x - a), \\ & y - b = \frac{\left( mm^{'} - 1 \right)}{\left( m + m^{'} \right)}(x - a) \end{matrix}$$

✔ (a) $y - b = \frac{m + m'}{1 - mm'}(x - a)$ is one bisector.

✘ (b) incorrect grouping of the slope terms.

✘ (c) incorrect grouping of the slope terms.

✔ (d) $y - b = \frac{mm' - 1}{m + m'}(x - a)$ is the perpendicular bisector.