#### Answer

$m_{1}=\frac{m_{2}-\tan a}{1+m_{2} \tan a}$

#### Work Step by Step

From the given figure:
$a_{1}=a_{2}-a$
$m_{1}$ is the slope of trajactory desired
$m_{2}$ is the slope of trajactory given
$m_{1}=\tan{a_{1}}$
and $m_{2}=\tan{a_{2}}$
Now,
$\tan{a_{1}}=\tan{(a_{2}-a)}$
Thus,
$m_{1}=\frac{\tan a_{2}-\tan a}{1+\tan{a_{2}\tan a}}$
$m_{1}=\frac{m_{2}-\tan a}{1+m_{2} \tan a}$
Thus, the final answer:
$m_{1}=\frac{m_{2}-\tan a}{1+m_{2} \tan a}$